Complete NCERT theory notes, Coulomb's Law vector derivations, electric field & dipole calculations, Gauss's law applications, solved examples 1.1–1.12, interactive quiz with stepwise explanations, and full exercises 1.1–1.23.
01 / Exam-Focused Notes
Chapter 4: Moving Charges and Magnetism
Comprehensive, exam-oriented study notes covering all 10 NCERT Sections (4.1 to 4.10), all 12 Solved Examples (4.1 to 4.12), complete step-by-step mathematical derivations, vector diagrams, and conceptual subtleties.
Oersted's Experiment (1820): Hans Christian Oersted noticed that an electric current flowing through a straight wire caused a noticeable deflection in a nearby magnetic compass needle.
The needle aligns tangential to concentric circles centered on the wire in a plane perpendicular to the wire.
Reversing the direction of current reverses the needle orientation.
Deflection increases with higher current $I$ and closer proximity $r$. Iron filings sprinkled on a cardboard sheet arrange in concentric circles.
Conclusion: Moving electric charges / currents produce a magnetic field $\mathbf{B}$ in surrounding space.
Dot and Cross Direction Conventions:
Dot ($\odot$): Current or field emerging normally out of the plane of paper (like the tip of an arrow approaching you).
Cross ($\otimes$): Current or field going normally into the plane of paper (like the feathered tail of an arrow moving away).
Core Derivation & Physics
4.2.1 Magnetic Field $\mathbf{B}$, Lorentz Force & Force on Conductor
Sources and Fields: Static charges produce an electric field $\mathbf{E}(\mathbf{r})$. Moving charges or currents produce both an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}(\mathbf{r})$. Both fields satisfy the principle of superposition.
Lorentz Force Formulation: Total force $\mathbf{F}$ on charge $q$ moving with velocity $\mathbf{v}$ in presence of both $\mathbf{E}$ and $\mathbf{B}$:
Salient Features of Magnetic Force ($\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$):
Magnitude: $F_m = |q| v B \sin\theta$, where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{B}$.
Direction: Perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ ($\mathbf{F}_m \perp \mathbf{v}$ and $\mathbf{F}_m \perp \mathbf{B}$) given by Right-Hand Screw Rule. Force on negative charge is opposite to that on positive charge.
Zero Force Conditions: $F_m = 0$ if particle is stationary ($v = 0$), or if velocity is parallel ($\theta = 0^\circ$) or antiparallel ($\theta = 180^\circ$) to $\mathbf{B}$.
Zero Work Done: Since $\mathbf{F}_m \perp \mathbf{v}$, power $P = \mathbf{F}_m \cdot \mathbf{v} = 0$. Hence work done by magnetic force on a charged particle is identically zero ($W = 0$). Magnetic force alters the direction of motion but cannot change the speed or kinetic energy.
SI Unit of Magnetic Field: Tesla ($\text{T} \equiv \text{N s C}^{-1}\text{ m}^{-1} \equiv \text{N A}^{-1}\text{ m}^{-1}$). Dimensions: $[M T^{-2} A^{-1}]$.
Definition: $B = 1\text{ T}$ if a charge of $1\text{ C}$ moving at $1\text{ m s}^{-1}$ perpendicular to the field experiences a magnetic force of $1\text{ N}$. Non-SI unit: Gauss ($1\text{ G} = 10^{-4}\text{ T}$). Earth's field $\approx 3.6 \times 10^{-5}\text{ T}$.
Magnetic Force on a Current-Carrying Conductor (Step-by-Step Derivation): Consider a straight rod of uniform cross-section $A$, length $l$, and electron number density $n$. Total mobile carriers $N = n A l$.
Each carrier moves with average drift velocity $\mathbf{v}_d$. Force on all carriers:
$\mathbf{F} = (n A l) q (\mathbf{v}_d \times \mathbf{B}) = [(n q \mathbf{v}_d) A l] \times \mathbf{B} = [\mathbf{j} A l] \times \mathbf{B}$
Transferring the vector direction from current density $\mathbf{j}$ to length vector $\mathbf{l}$ (in direction of current $I = j A$):
$\mathbf{F} = I (\mathbf{l} \times \mathbf{B}) = I l B \sin\theta\,\hat{\mathbf{n}}$
For an arbitrarily curved wire: $\mathbf{F} = \int I (d\mathbf{l} \times \mathbf{B})$.
Example 4.1
Mid-Air Suspension of Current-Carrying Wire
A straight wire of mass $200\text{ g}$ and length $1.5\text{ m}$ carries a current of $2\text{ A}$. It is suspended in mid-air by a uniform horizontal magnetic field $\mathbf{B}$ (Fig. 4.3). What is the magnitude of the magnetic field?
Equilibrium Condition: For vertical mid-air suspension, upward magnetic force $F = I l B$ must exactly balance downward gravitational force $F_g = m g$:
$I l B = m g \implies B = \frac{m g}{I l}$
Direction of Lorentz Force on Moving Electron and Proton
If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge)?
Vector Cross Product: Velocity $\mathbf{v} = v\,\hat{\mathbf{i}}$, Magnetic field $\mathbf{B} = B\,\hat{\mathbf{j}}$.
$\mathbf{v} \times \mathbf{B} = (v\,\hat{\mathbf{i}}) \times (B\,\hat{\mathbf{j}}) = v B\,\hat{\mathbf{k}} \quad (\text{along } +z\text{-axis})$
(a) For Electron ($q = -e$): $\mathbf{F} = -e(\mathbf{v} \times \mathbf{B}) = -e v B\,\hat{\mathbf{k}} \implies$ directed along $-z$-axis (into page).
(b) For Proton ($q = +e$): $\mathbf{F} = +e(\mathbf{v} \times \mathbf{B}) = +e v B\,\hat{\mathbf{k}} \implies$ directed along $+z$-axis (out of page).
Answers: (a) Along $-z$ axis, (b) Along $+z$ axis.
4.3
NCERT Section
Motion in a Magnetic Field & Helical Trajectories
Kinematics in B-Field
4.3.1 Circular Motion, Cyclotron Frequency & Pitch of Helix
Case I: Velocity Perpendicular to Field ($\mathbf{v} \perp \mathbf{B}$, $\theta = 90^\circ$): Magnetic force $F = q v B$ acts perpendicular to $\mathbf{v}$ towards the center of orbit, providing the required centripetal force:
$\frac{m v^2}{r} = q v B$
$\text{Orbit Radius: } \quad r = \frac{m v}{q B} = \frac{p}{q B} = \frac{\sqrt{2 m K}}{q B}$
Angular Frequency ($\omega$), Cyclotron Frequency ($\nu$) & Time Period ($T$):
$\omega = \frac{v}{r} = \frac{q B}{m}$
Fundamental Property: Both rotation frequency $\nu$ and period $T$ are independent of velocity $v$, orbital radius $r$, and energy. Fast particles describe larger circles, completing one revolution in exactly the same time.
Case II: Velocity at Arbitrary Angle $\theta$ to $\mathbf{B}$ (Helical Motion):
Parallel velocity component $v_\parallel = v\cos\theta$ is unaffected by $\mathbf{B}$ ($F_\parallel = 0$), causing steady translation along $\mathbf{B}$.
Resulting path is a helix with axis along $\mathbf{B}$.
Pitch of Helix ($p$): Linear distance moved along magnetic field during one complete revolution:
$p = v_\parallel T = v\cos\theta \left(\frac{2\pi m}{q B}\right) = \frac{2\pi m v \cos\theta}{q B}$
Example 4.3
Radius, Frequency and Kinetic Energy of Orbiting Electron
What is the radius of the path of an electron ($m = 9 \times 10^{-31}\text{ kg}, q = 1.6 \times 10^{-19}\text{ C}$) moving at speed $3 \times 10^7\text{ m s}^{-1}$ in a magnetic field of $6 \times 10^{-4}\text{ T}$ perpendicular to it? What is its frequency? Calculate its energy in keV.
4.4.1 Biot-Savart Law Statement, Vector Form & Coulomb Analogy
The magnetic field $d\mathbf{B}$ due to an infinitesimal current element $I d\mathbf{l}$ at distance $r$ and angle $\theta$ to displacement $\mathbf{r}$:
Proportionality Constant & Permeability ($\mu_0$):
$\frac{\mu_0}{4\pi} = 10^{-7}\text{ T m A}^{-1} \implies \mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1} \quad (\text{or }\text{N A}^{-2})$
$\mu_0$ is called the permeability of free space (vacuum). Dimensions: $[M L T^{-2} A^{-2}]$.
Connection to Permittivity ($\varepsilon_0$) and Speed of Light ($c$):
$(\varepsilon_0 \mu_0) = \left(\frac{1}{4\pi \times 9 \times 10^9}\right) \times (4\pi \times 10^{-7}) = \frac{10^{-7}}{9 \times 10^9} = \frac{1}{(3 \times 10^8)^2} = \frac{1}{c^2} \implies c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
Comparison Between Biot-Savart Law & Coulomb's Law:
Feature
Coulomb's Law (Electrostatics)
Biot-Savart Law (Magnetism)
Source
Scalar charge $dq$
Vector current element $I d\mathbf{l}$
Direction
Along displacement vector $\mathbf{r}$ (radial)
Perpendicular to both $d\mathbf{l}$ and $\mathbf{r}$ ($d\mathbf{l} \times \mathbf{r}$)
Angle Dependence
Independent of orientation (isotropic)
Depends on $\sin\theta$; field is zero along axis of element ($\theta = 0$)
Range
Inverse-square law ($1/r^2$), long range
Inverse-square law ($1/r^2$), long range
Superposition
Obeyed vectorially
Obeyed vectorially
Example 4.4
Magnetic Field of Current Element on Y-Axis
An element $\Delta\mathbf{l} = \Delta x\,\hat{\mathbf{i}}$ is placed at the origin and carries current $I = 10\text{ A}$ (Fig. 4.8). What is the magnetic field on the y-axis at distance $y = 0.5\text{ m}$, given $\Delta x = 1\text{ cm}$?
Answer: $\mathbf{B} = 4 \times 10^{-8}\text{ T}\,\hat{\mathbf{k}}$ (along $+z$-axis, out of page).
4.5
NCERT Section
Magnetic Field on the Axis of a Circular Current Loop
Complete Master Derivation
4.5.1 Axial Field Derivation & Centre Special Cases
Geometry: Circular loop of radius $R$ in $y$-$z$ plane carrying steady current $I$. Point $P$ on $x$-axis at distance $x$ from center $O$.
Biot-Savart on Element $d\mathbf{l}$ of Loop: Distance to point $P$ is $r = \sqrt{x^2 + R^2}$.
Since element $d\mathbf{l}$ is in $y$-$z$ plane and displacement $\mathbf{r}$ has components in $x$ and $y/z$, $d\mathbf{l} \perp \mathbf{r} \implies |d\mathbf{l} \times \mathbf{r}| = r dl$.
$dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl}{x^2 + R^2}$
Resolution into Components & Cancellation:
Perpendicular components $dB_\perp = dB\sin\theta$ due to diametrically opposite pairs cancel exactly ($\int dB_\perp = 0$).
Axial components $dB_x = dB\cos\theta$ add constructively along $+x$ axis, where $\cos\theta = \frac{R}{r} = \frac{R}{\sqrt{x^2 + R^2}}$.
Integration over Full Circumference ($\oint dl = 2\pi R$):
$B_x = \int dB_x = \int \frac{\mu_0 I dl}{4\pi(x^2 + R^2)} \frac{R}{(x^2 + R^2)^{1/2}} = \frac{\mu_0 I R}{4\pi(x^2 + R^2)^{3/2}} \oint dl = \frac{\mu_0 I R (2\pi R)}{4\pi(x^2 + R^2)^{3/2}}$
$\mathbf{B}_{\text{axial}}(x) = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}\,\hat{\mathbf{i}} \qquad \left(\text{For } N \text{ turns: } B(x) = \frac{\mu_0 N I R^2}{2(x^2 + R^2)^{3/2}}\right)$
Special Case 1: Field at the Centre of the Loop ($x = 0$):
$B_{\text{centre}} = \frac{\mu_0 I}{2 R} \qquad \left(\text{For } N \text{ turns: } B_{\text{centre}} = \frac{\mu_0 N I}{2 R}\right)$
Special Case 2: Semicircular Arc of Radius $R$: $B_{\text{semi}} = \frac{1}{2} B_{\text{centre}} = \frac{\mu_0 I}{4 R}$.
For arc subtending angle $\phi$ (in radians) at centre: $B_{\text{arc}} = \frac{\mu_0 I \phi}{4\pi R}$.
Special Case 3: Far Axial Distance ($x \gg R$):
$B \approx \frac{\mu_0 I R^2}{2 x^3} = \frac{\mu_0 (I \pi R^2)}{2\pi x^3} = \frac{\mu_0 I A}{2\pi x^3} = \frac{\mu_0}{4\pi} \frac{2 m}{x^3} \quad (\text{Dipole Field})$
Right-Hand Thumb Rule for Loop: Curl fingers of right hand along the circular current; the extended thumb points in the direction of the axial magnetic field lines.
Fig 4.11: Magnetic Field along the Axis of a Circular Current Loop (Biot-Savart Law)
Example 4.5
Magnetic Field at Centre of Semicircular Arc
A straight wire carrying current $12\text{ A}$ is bent into a semicircular arc of radius $2.0\text{ cm}$ (Fig. 4.11). (a) Field due to straight segments? (b) Semicircle contribution? (c) Bent opposite way?
(a) Straight Segments: For all elements along the straight segments, $d\mathbf{l} \parallel \mathbf{r} \implies d\mathbf{l} \times \mathbf{r} = \mathbf{0} \implies B_{\text{straight}} = 0$.
(b) Semicircular Arc: $d\mathbf{l} \perp \mathbf{r}$ everywhere. Field is half of complete circle:
$B = \frac{\mu_0 I}{4 R} = \frac{(4\pi \times 10^{-7}) \times 12}{4 \times 0.02} = 6\pi \times 10^{-5}\text{ T} \approx 1.9 \times 10^{-4}\text{ T} \quad (\text{directed into paper})$
(c) Inverted Semicircle: Magnitude is identical ($1.9 \times 10^{-4}\text{ T}$), but directed out of the plane of paper.
Answers: (a) $0$, (b) $1.9 \times 10^{-4}\text{ T}$ into page, (c) $1.9 \times 10^{-4}\text{ T}$ out of page.
Example 4.6
Field at Centre of 100-Turn Circular Coil
Consider a tightly wound 100-turn coil of radius $10\text{ cm}$ carrying current $1\text{ A}$. What is the magnetic field at the centre?
4.6.1 Ampere's Law Statement, Cylindrical Wire & Solenoid Field
Ampere's Circuital Law: The line integral of magnetic field $\mathbf{B}$ around any closed boundary loop $C$ (Amperian loop) equals $\mu_0$ times the total steady current $I_e$ passing through the open surface bounded by $C$:
Sign Convention: Curl right-hand fingers in the direction of traversal along boundary $C$; the extended thumb gives the positive direction for current $I_e$.
Simplified Form for Symmetric Systems ($B$ tangential and constant along length $L$):
$B L = \mu_0 I_e$
Application 1: Long Straight Infinite Wire: Circular Amperian loop of radius $r$: $\oint \mathbf{B}\cdot d\mathbf{l} = B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$.
Application 2: Thick Wire of Radius $a$ with Uniform Current Density:
Outside Wire ($r \ge a$): $I_{\text{enc}} = I \implies B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$.
Inside Wire ($r < a$): $I_{\text{enc}} = I \frac{\pi r^2}{\pi a^2} = I \frac{r^2}{a^2} \implies B(2\pi r) = \mu_0 I \frac{r^2}{a^2} \implies B = \left(\frac{\mu_0 I}{2\pi a^2}\right) r \propto r$.
At the boundary ($r = a$): $B_{\text{max}} = \frac{\mu_0 I}{2\pi a}$.
4.7 The Long Solenoid ($L \gg R$):
Helically wound closely spaced insulated turns ($n = N/L$ turns per unit length).
Inside: Field is strong, uniform, and parallel to axis. Outside: Field approaches zero ($B_{\text{out}} \approx 0$).
Amperian Rectangular Loop $abcd$ (Length $h$):
$\oint \mathbf{B}\cdot d\mathbf{l} = \int_a^b \mathbf{B}\cdot d\mathbf{l} + \int_b^c \mathbf{B}\cdot d\mathbf{l} + \int_c^d \mathbf{B}\cdot d\mathbf{l} + \int_d^a \mathbf{B}\cdot d\mathbf{l} = B h + 0 + 0 + 0 = B h$
Current enclosed: $I_e = n h I$.
$B h = \mu_0 (n h I) \implies B = \mu_0 n I = \mu_0 \left(\frac{N}{L}\right) I$
Field at Ends of Finite Solenoid: $B_{\text{end}} = \frac{1}{2} \mu_0 n I$.
Example 4.7
Field Inside and Outside a Solid Cylindrical Conductor
A long straight wire of circular cross-section (radius $a$) carries steady current $I$ uniformly distributed. Calculate $B$ in regions $r < a$ and $r > a$.
(a) Outside ($r > a$): Amperian circle $L = 2\pi r$, $I_e = I \implies B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$.
(b) Inside ($r < a$): Current density $j = \frac{I}{\pi a^2}$. Current enclosed $I_e = j(\pi r^2) = \frac{I r^2}{a^2}$.
$B(2\pi r) = \mu_0 \left(\frac{I r^2}{a^2}\right) \implies B = \frac{\mu_0 I r}{2\pi a^2} \propto r$
Answers: For $r \ge a$: $B = \frac{\mu_0 I}{2\pi r}$; For $r \le a$: $B = \frac{\mu_0 I r}{2\pi a^2}$.
Example 4.8
Magnetic Field Inside a Long Solenoid
A solenoid of length $0.5\text{ m}$ has radius $1\text{ cm}$ and is made up of 500 turns. It carries a current of $5\text{ A}$. What is the magnitude of the magnetic field inside the solenoid?
Turns per unit length ($n$): $n = \frac{N}{L} = \frac{500}{0.5\text{ m}} = 1000\text{ turns/m}$. Since $L/a = 0.5/0.01 = 50 \gg 1$, long solenoid formula applies.
Calculation:
$B = \mu_0 n I = (4\pi \times 10^{-7}\text{ T m A}^{-1}) \times 1000\text{ m}^{-1} \times 5\text{ A} = 2\pi \times 10^{-3}\text{ T} \approx 6.28 \times 10^{-3}\text{ T}$
Answer: $B = 6.28 \times 10^{-3}\text{ T}$.
4.8
NCERT Section
Force Between Parallel Conductors & The Ampere
Mutual Force Derivation
4.8.1 Parallel Wires Force & Definition of the SI Ampere
Derivation: Two long parallel conductors $a$ and $b$ separated by distance $d$ carrying steady currents $I_a, I_b$:
Conductor $a$ produces magnetic field at conductor $b$: $B_a = \frac{\mu_0 I_a}{2\pi d}$ (directed downwards into plane).
Conductor $b$ experiences force on segment of length $L$: $F_{ba} = I_b L B_a = \frac{\mu_0 I_a I_b L}{2\pi d}$ (directed towards $a$).
Similarly, force on $a$ due to $b$ is $F_{ab} = \frac{\mu_0 I_a I_b L}{2\pi d}$ (directed towards $b$).
$\mathbf{F}_{ba} = -\mathbf{F}_{ab} \quad (\text{Consistent with Newton's Third Law})$
$\text{Force per Unit Length: } \quad f = \frac{F}{L} = \frac{\mu_0 I_a I_b}{2\pi d} \quad (\text{SI Unit: }\text{N m}^{-1})$
Attraction vs Repulsion Rule:
Parallel Currents (Same Direction): ATTRACT each other.
Antiparallel Currents (Opposite Direction): REPEL each other.
Contrast with Electrostatics: Like charges repel, but like (parallel) currents attract!
Standard SI Definition of 1 Ampere (1946): The ampere is that steady current which, when maintained in each of two infinitely long, straight, parallel conductors of negligible cross-section, placed $1\text{ metre}$ apart in vacuum, produces on each conductor a force equal to $2 \times 10^{-7}\text{ newtons per metre of length}$.
Definition of 1 Coulomb: Charge transported through a cross-section by steady current of $1\text{ A}$ in $1\text{ s}$ ($1\text{ C} = 1\text{ A s}$).
Example 4.9
Earth's Magnetic Force on Current-Carrying Conductor
Horizontal component of Earth's field is $B_H = 3.0 \times 10^{-5}\text{ T}$ from South to North. A conductor carries steady current $1\text{ A}$. Find force per unit length when current flows: (a) East to West, (b) South to North.
Formula: $f = \frac{F}{l} = I B \sin\theta$.
(a) Current East to West: $\theta = 90^\circ \implies f = I B_H \sin 90^\circ = 1\text{ A} \times 3.0 \times 10^{-5}\text{ T} = 3.0 \times 10^{-5}\text{ N m}^{-1}$. By right-hand cross product $\mathbf{l} \times \mathbf{B}$ (West $\times$ North), direction is vertically downwards.
(b) Current South to North: $\theta = 0^\circ \implies f = I B_H \sin 0^\circ = 0$. (No force on conductor).
Answers: (a) $3.0 \times 10^{-5}\text{ N m}^{-1}$ downwards, (b) $0$.
4.9
NCERT Section
Torque on Current Loop & Circular Loop as Magnetic Dipole
Torque Derivation & Dipole Moment
4.9.1 Torque on Planar Current Loop ($\boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}$)
Rectangular Loop $ABCD$ in Uniform Field $\mathbf{B}$ (Sides $a$ and $b$, Area $A = ab$):
Forces on arms $BC$ and $DA$ are equal, opposite, and collinear along the loop axis $\implies$ cancel out completely ($F = 0, \tau = 0$).
Forces on arms $AB$ and $CD$ have magnitude $F_1 = F_2 = I b B$. They are equal and opposite, but not collinear, forming a deflecting couple.
Perpendicular lever arm between the two forces is $a\sin\theta$, where $\theta$ is the angle between normal/area vector $\mathbf{A}$ and magnetic field $\mathbf{B}$.
$\tau = F_1 \left(\frac{a}{2}\sin\theta\right) + F_2 \left(\frac{a}{2}\sin\theta\right) = (I b B) a \sin\theta = I (ab) B \sin\theta = I A B \sin\theta$
Magnetic Dipole Moment ($\mathbf{m}$): $\mathbf{m} = I \mathbf{A} \quad (\text{For } N \text{ turns: } \mathbf{m} = N I \mathbf{A})$. SI Unit: $\text{A m}^2 \equiv \text{J T}^{-1}$. Dimensions: $[L^2 A]$.
Direction of $\mathbf{m}$ is given by right-hand rule (curl fingers along current, thumb points along $\mathbf{m}$).
4.9.2 Circular Current Loop as Magnetic Dipole (Analogy with Electrostatic Dipole): Axial field of circular loop at large distance $x \gg R$:
$B_{\text{axial}} = \frac{\mu_0 I R^2}{2 x^3} = \frac{\mu_0 (I \pi R^2)}{2\pi x^3} = \frac{\mu_0 m}{2\pi x^3} = \frac{\mu_0}{4\pi} \frac{2 m}{x^3}$
Equatorial field in plane of loop at $x \gg R$:
$B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{m}{x^3}$
Exact mathematical correspondence with electric dipole: $\mathbf{E} \leftrightarrow \mathbf{B}$, $\mathbf{p} \leftrightarrow \mathbf{m}$, $\frac{1}{\varepsilon_0} \leftrightarrow \mu_0$.
Key difference: Electric dipoles have isolated monopoles (charges $\pm q$), whereas magnetic monopoles do not exist in nature.
Example 4.10
Field, Moment, Torques and Angular Speed of Rotating Coil
A 100-turn circular coil of radius $10\text{ cm}$ carries $3.2\text{ A}$. (a) $B$ at centre? (b) Magnetic moment $m$? (c) Torques at initial ($\theta = 0^\circ$) and final ($\theta = 90^\circ$) positions in horizontal $B = 2\text{ T}$? (d) Angular speed $\omega$ acquired after rotating $90^\circ$, given moment of inertia $\mathcal{I} = 0.1\text{ kg m}^2$?
(a) Field at Centre: $B = \frac{\mu_0 N I}{2 R} = \frac{(4\pi \times 10^{-7}) \times 100 \times 3.2}{2 \times 0.1} \approx 2 \times 10^{-3}\text{ T}$.
(b) Magnetic Moment: $m = N I A = 100 \times 3.2 \times (\pi \times 0.1^2) = 320 \times 0.0314 \approx 10\text{ A m}^2$.
(c) Torques: Initial ($\theta = 0^\circ$): $\tau_i = m B \sin 0^\circ = 0$. Final ($\theta = 90^\circ$): $\tau_f = m B \sin 90^\circ = 10 \times 2 = 20\text{ N m}$.
(d) Angular Speed ($\omega$): $\tau = \mathcal{I}\alpha = \mathcal{I} \frac{d\omega}{dt} = \mathcal{I}\omega \frac{d\omega}{d\theta} = m B \sin\theta$.
$\int_0^\omega \mathcal{I}\omega\,d\omega = \int_0^{\pi/2} m B \sin\theta\,d\theta \implies \frac{1}{2}\mathcal{I}\omega^2 = m B [-\cos\theta]_0^{\pi/2} = m B$
$\omega = \sqrt{\frac{2 m B}{\mathcal{I}}} = \sqrt{\frac{2 \times 10 \times 2}{0.1}} = \sqrt{400} = 20\text{ rad s}^{-1}$
Answers: (a) $2 \times 10^{-3}\text{ T}$, (b) $10\text{ A m}^2$, (c) $\tau_i = 0, \tau_f = 20\text{ N m}$, (d) $\omega = 20\text{ rad s}^{-1}$.
Example 4.11
Conceptual Questions on Current Loop in Magnetic Field
(a) Can a horizontal current loop turn around vertical axis in uniform $\mathbf{B}$? (b) Stable equilibrium orientation and flux? (c) Why does a flexible irregular loop become circular in $\mathbf{B}$?
(a) Vertical Rotation: No, because turning about vertical axis requires torque $\boldsymbol{\tau}$ to be vertical. But $\boldsymbol{\tau} = I(\mathbf{A} \times \mathbf{B})$ is always in the horizontal plane because area vector $\mathbf{A}$ is vertical.
(b) Stable Orientation & Flux: Stable equilibrium occurs when $\mathbf{A}$ is aligned with external $\mathbf{B}$ ($\theta = 0^\circ$). In this state, field produced by loop is parallel to external field, maximizing total magnetic flux $\Phi$.
(c) Circular Shape: To maximize enclosed magnetic flux $\Phi = B A$, the flexible loop expands into a circular shape because for a given fixed perimeter, a circle encloses maximum possible area.
Answers: (a) No, (b) $\mathbf{A} \parallel \mathbf{B}$ (Max flux), (c) Maximizes area and magnetic flux.
4.10
NCERT Section
The Moving Coil Galvanometer (MCG)
Instrument Physics & Conversions
4.10.1 Principle, Radial Magnetic Field, Sensitivities & Meter Conversions
Principle: When a current $I$ flows through a coil placed in a magnetic field, it experiences a deflecting magnetic torque:
$\tau = N I A B \sin\theta$
Role of Soft Iron Cylindrical Core & Concave Pole Pieces:
Makes the magnetic field radial, ensuring the plane of coil is always parallel to field lines ($\theta = 90^\circ \implies \sin\theta = 1$) in all deflection positions.
High magnetic permeability $\mu_r$ of soft iron significantly intensifies the magnetic field strength $B$.
Equilibrium Deflection Equation: Deflecting torque is balanced by the restoring torque of phosphor-bronze spring ($k\phi$):
$N I A B = k \phi \implies \phi = \left(\frac{N A B}{k}\right) I \implies \phi \propto I$
where $k$ is the torsional constant (restoring torque per unit twist) of the suspension spring. The linear relation $\phi \propto I$ gives a uniform linear scale.
Current Sensitivity ($S_i$): Deflection produced per unit current:
$S_i = \frac{\phi}{I} = \frac{N A B}{k} \quad (\text{SI Unit: }\text{rad A}^{-1} \text{ or div/A})$
Voltage Sensitivity ($S_v$): Deflection produced per unit applied voltage:
$S_v = \frac{\phi}{V} = \frac{\phi}{I R_G} = \frac{N A B}{k R_G} \quad (\text{SI Unit: }\text{rad V}^{-1} \text{ or div/V})$
Important NCERT Subtlety: Why increasing Current Sensitivity may not increase Voltage Sensitivity?
If number of turns is doubled ($N \to 2N$), current sensitivity doubles ($S_i \to 2S_i$). However, doubling turns doubles the wire length and hence coil resistance ($R_G \to 2R_G$). Thus $S_v = \frac{2 N A B}{k (2 R_G)} = S_v$ remains unchanged.
Conversion of Galvanometer to Ammeter: Galvanometer has high resistance $R_G$ and small full-scale deflection current $I_g$ ($\sim \mu\text{A}$). To convert into ammeter of range $I$ ($I > I_g$), connect a small shunt resistance $r_s$ in parallel:
$V_{\text{galv}} = V_{\text{shunt}} \implies I_g R_G = (I - I_g) r_s$
Conversion of Galvanometer to Voltmeter: To measure voltage across range $V$ without drawing significant circuit current, connect a high multiplier resistance $R$ in series:
$V = I_g(R_G + R)$
$R = \frac{V}{I_g} - R_G, \qquad R_V = R_G + R \approx R \quad (\text{Ideal Voltmeter: } R_V = \infty)$
Circuit Current Measurement with Galvanometer, Ammeter & Ideal Meter
In a circuit with $3.0\text{ V}$ battery and $3.0\,\Omega$ resistor, what is current if ammeter is: (a) Galvanometer with $R_G = 60.00\,\Omega$, (b) Galvanometer converted with shunt $r_s = 0.02\,\Omega$, (c) Ideal ammeter with $R = 0$?
Work done by a magnetic force on a moving charged particle is always:
Option A is correct. Magnetic force $\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$ is always perpendicular to velocity $\mathbf{v}$ ($\mathbf{F}_m \cdot \mathbf{v} = 0$). Hence power $P = \mathbf{F}_m \cdot \mathbf{v} = 0$ and work done $W = 0$.
Option B is incorrect. Electric force can do work, but magnetic force cannot.
Option C is incorrect. Dot product with displacement is identically zero.
Option D is incorrect. Kinetic energy remains constant throughout.
Core Rule
Magnetic force changes the direction of motion, never the speed or kinetic energy ($W = 0$).
Q2UNANSWERED
The cyclotron frequency of a charged particle of mass $m$ and charge $q$ in magnetic field $B$ is independent of:
Option A is correct. $\nu_c = \frac{q B}{2\pi m}$, which depends strictly on $q/m$ and $B$, completely independent of velocity $v$, energy, or orbital radius $r$.
Option B is incorrect. $\nu_c$ is directly proportional to $B$.
Option C is incorrect. $\nu_c$ is proportional to charge $q$.
Option D is incorrect. $\nu_c$ is inversely proportional to mass $m$.
Core Rule
Cyclotron resonance condition: Period $T = \frac{2\pi m}{q B}$ is invariant with particle energy.
Q3UNANSWERED
Two long parallel wires carrying currents in the same direction:
Option A is correct. By right-hand thumb rule and Fleming's left-hand rule, parallel currents attract with force per unit length $f = \frac{\mu_0 I_1 I_2}{2\pi d}$.
Option B is incorrect. Antiparallel currents repel; parallel currents attract.
Option C is incorrect. Mutual magnetic force is non-zero.
Option D is incorrect. Parallel wires experience pure attractive translation.
To convert a galvanometer into an ammeter of range $I$, we connect:
Option A is correct. A low resistance shunt $r_s = \frac{I_g R_G}{I - I_g}$ in parallel bypasses excess current and ensures overall ammeter resistance is negligible.
Option B is incorrect. High resistance in series is used to convert galvanometer to a voltmeter.
Option C is incorrect. Shunts must be connected in parallel.
Option D is incorrect. Parallel resistance must be very small.
Core Rule
Ammeter $\to$ Small shunt in parallel ($r_s \ll R_G$). Voltmeter $\to$ Large multiplier in series ($R \gg R_G$).
Q5UNANSWERED
If the number of turns $N$ of a galvanometer coil is doubled ($N \to 2N$), its voltage sensitivity will:
Option A is correct. $S_v = \frac{N A B}{k R_G}$. When $N \to 2N$, wire length doubles so resistance also doubles ($R_G \to 2R_G$). Thus $S_v$ remains invariant even though current sensitivity doubles.
Option B is incorrect. Current sensitivity doubles, but voltage sensitivity does not.
Option C is incorrect. The factor of 2 cancels out.
Option D is incorrect. $S_v$ is invariant.
Core Rule
Increasing current sensitivity by adding turns does not increase voltage sensitivity because coil resistance increases proportionally.
03 / Textbook Solutions
NCERT Exercises 4.1 – 4.13
Complete stepwise solutions for every textbook exercise question in NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism Reprint 2026-27.
13
QUESTIONS
Ex 4.1Centre of Circular Coil
4.1 A circular coil of wire consisting of 100 turns, each of radius $8.0\text{ cm}$ carries a current of $0.40\text{ A}$. What is the magnitude of the magnetic field $\mathbf{B}$ at the centre of the coil?
4.3 A long straight wire in the horizontal plane carries a current of $50\text{ A}$ in north to south direction. Give the magnitude and direction of $\mathbf{B}$ at a point $2.5\text{ m}$ east of the wire.
Direction (Right-Hand Rule): Thumb pointing North $\to$ South, fingers curl out of horizontal plane at point east $\implies$ Vertically Upwards.
Ex 4.4Power Line Field
4.4 A horizontal overhead power line carries a current of $90\text{ A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5\text{ m}$ below the line?
$B = 1.2 \times 10^{-5}\text{ T}$, directed horizontally towards South.
Direction: Current flows East $\to$ West. By right-hand thumb rule, below the wire fingers point towards Geographic South.
Ex 4.5Force per Unit Length
4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of $8\text{ A}$ and making an angle of $30^\circ$ with the direction of a uniform magnetic field of $0.15\text{ T}$?
4.6 A $3.0\text{ cm}$ wire carrying a current of $10\text{ A}$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is $0.27\text{ T}$. What is the magnetic force on the wire?
$F = 8.1 \times 10^{-2}\text{ N}$.
Calculation: $\theta = 90^\circ \implies F = I l B = 10\text{ A} \times (3.0 \times 10^{-2}\text{ m}) \times 0.27\text{ T} = 8.1 \times 10^{-2}\text{ N}$.
Ex 4.7Parallel Wire Mutual Force
4.7 Two long and parallel straight wires A and B carrying currents of $8.0\text{ A}$ and $5.0\text{ A}$ in the same direction are separated by $4.0\text{ cm}$. Estimate the force on a $10\text{ cm}$ section of wire A.
$F = 2.0 \times 10^{-5}\text{ N}$ (Attractive force towards wire B).
4.8 A closely wound solenoid $80\text{ cm}$ long has 5 layers of windings of 400 turns each. Diameter is $1.8\text{ cm}$. If current is $8.0\text{ A}$, estimate $B$ inside near centre.
4.9 A square coil of side $10\text{ cm}$ consists of 20 turns and carries current of $12\text{ A}$. Suspended vertically, normal to coil makes $30^\circ$ with uniform horizontal $B = 0.80\text{ T}$. Find torque.
$\tau = 0.96\text{ N m}$.
Area: $A = (0.10)^2 = 0.01\text{ m}^2$.
Torque: $\tau = N I A B \sin\theta = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^\circ = 1.92 \times 0.5 = 0.96\text{ N m}$.
Ex 4.10Galvanometer Sensitivity Ratios
4.10 Two moving coil meters $M_1$ and $M_2$: ($R_1 = 10\,\Omega, N_1 = 30, A_1 = 3.6 \times 10^{-3}, B_1 = 0.25\text{ T}$) and ($R_2 = 14\,\Omega, N_2 = 42, A_2 = 1.8 \times 10^{-3}, B_2 = 0.50\text{ T}$). Identical spring constants. Find ratio of (a) Current sensitivity, (b) Voltage sensitivity of $M_2$ to $M_1$.
4.11 In a chamber, $B = 6.5\text{ G} = 6.5 \times 10^{-4}\text{ T}$. Electron shot with speed $4.8 \times 10^6\text{ m s}^{-1}$ normal to field. Explain why path is circle and find radius. ($e = 1.6 \times 10^{-19}\text{ C}, m = 9.1 \times 10^{-31}\text{ kg}$).
Path is circular as $\mathbf{F}_m \perp \mathbf{v}$ acts as centripetal force • $r = 4.2\text{ cm}$.
4.13 (a) Circular coil of 30 turns and radius $8.0\text{ cm}$ carrying current $6.0\text{ A}$ is suspended vertically in uniform horizontal $B = 1.0\text{ T}$. Field lines make $60^\circ$ with normal. Find counter torque. (b) Would answer change for planar coil of irregular shape enclosing same area?
(a) $\tau = 3.13\text{ N m}$ • (b) No, torque depends solely on area $A$, not shape.
(a) Torque: $A = \pi R^2 = \pi \times (0.08)^2 \approx 0.0201\text{ m}^2$.
$\tau = N I A B \sin 60^\circ = 30 \times 6.0 \times 0.0201 \times 1.0 \times \frac{\sqrt{3}}{2} \approx 3.13\text{ N m}$.
(b) Irregular Shape: Torque $\boldsymbol{\tau} = I \mathbf{A} \times \mathbf{B}$ depends only on the total enclosed area vector $\mathbf{A}$ and is strictly independent of the geometric perimeter shape.
04 / Rapid Reference
Chapter Summary & Points to Ponder
Core master principles, physical quantities table with SI units and dimensions, and all official NCERT Points to Ponder for Chapter 4.
100%
SYLLABUS
Lorentz Force Law
F = q(E + v × B) • W_m = 0
Magnetic force does zero work. Speed and KE are conserved.
Motion in B-Field
r = mv/qB • ν_c = qB/2πm
Cyclotron frequency $\nu_c$ is independent of speed and radius.
Biot-Savart Law
dB = (μ₀/4π) I dl×r / r³
Centre of circular loop: $B = \mu_0 I / 2R$. Axis: $B = \frac{\mu_0 I R^2}{2(x^2+R^2)^{3/2}}$.
Ampere's Circuital Law
∮ B·dl = μ₀ I_enc
Long wire: $B = \mu_0 I / 2\pi r$. Solenoid: $B = \mu_0 n I$.
Parallel Wire Force
f = μ₀ I₁I₂ / 2πd
Parallel attract; antiparallel repel. Defines the SI Ampere.
Torque on Dipole
τ = m × B = NIAB sinθ
Magnetic dipole moment $\mathbf{m} = N I \mathbf{A}$.
Moving Coil Galvanometer
φ = (NAB/k) I
Radial field from soft iron core ensures linear deflecting scale.
Continuous Closed Loops: Electrostatic field lines originate on positive charges and terminate on negative charges. Magnetic field lines always form continuous closed loops without start or end points (reflecting non-existence of magnetic monopoles).
Momentum of Electromagnetic Fields: In steady current circuits, Newton's third law holds directly for mutual forces ($\mathbf{F}_{ba} = -\mathbf{F}_{ab}$). For time-varying fields, conservation of momentum holds only when the momentum carried by electromagnetic fields is included.
Lorentz Force & Relativity: The magnetic force $\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$ depends on particle velocity $\mathbf{v}$. If we switch to a reference frame moving with velocity $\mathbf{v}$, the magnetic force vanishes and the acceleration is explained entirely by an electric field in that frame, showing electricity and magnetism are linked relativistic phenomena.
Ampere's vs Biot-Savart Law: Ampere's circuital law is not independent of Biot-Savart law; it can be derived from it. Their relationship is identical to that between Gauss's law and Coulomb's law in electrostatics.
Intrinsic Dipole Moments: Elementary particles like electrons and protons possess intrinsic magnetic dipole moments and spin, beyond circulating currents.
05 / CBSE 10-Year PYQ Bank
Chapter 4: Previous Year Questions
Complete 105 official CBSE Board & Sample Paper questions (2015–2026). Sorted topic-wise with step-wise board marking scheme solutions.
105
QUESTIONS
TOPIC 1
Concept of Magnetic Field & Oersted's Experiment
7 Questions
PYQ 1MCQ — 1M | CBSE Recurring 2016–2026
Topic 1
1. Oersted's experiment demonstrated that:
(a) A stationary charge creates a magnetic field
(b) A moving charge (electric current) creates a magnetic field
(c) A magnetic field creates an electric current
(d) Electric and magnetic fields are independent of each other
Correct Answer: (b) A moving charge (electric current) creates a magnetic field
Step 1: Oersted's Experiment observation
When an electric current flows through a wire, a compass needle placed nearby deflects. Since a compass needle only deflects in the presence of a magnetic field, the electric current must have produced a magnetic field around the wire.
Step 2: Conclusion
Since current is the flow of moving charges, this experiment proved that a moving charge (electric current) creates a magnetic field. Correct option is (b).
PYQ 2MCQ — 1M | CBSE Recurring
Topic 1
2. The SI unit of magnetic field intensity (magnetic flux density) is:
(a) Gauss
(b) Weber
(c) Tesla
(d) Henry
Correct Answer: (c) Tesla
Step 1: SI Unit of Magnetic Field
The magnetic field strength $B$ is measured in Tesla (T) in the SI system. Note: Gauss is the CGS unit ($1\text{ Tesla} = 10^4\text{ Gauss}$), Weber is the SI unit of magnetic flux, and Henry is the unit of inductance. Correct option is (c).
PYQ 3MCQ — 1M | CBSE Recurring 2018–2026
Topic 1
3. A compass needle placed near a current-carrying wire deflects. When the current in the wire is reversed, the deflection of the needle:
(a) Becomes zero
(b) Doubles
(c) Reverses direction
(d) Remains the same
Correct Answer: (c) Reverses direction
Step 1: Right-hand rule application
The direction of the magnetic field due to a straight wire carrying current is given by the right-hand thumb rule. If the direction of the current is reversed, the direction of the magnetic field lines reverses (e.g. from clockwise to counter-clockwise).
Step 2: Effect on Compass Needle
Since the magnetic field direction is reversed, the force on the compass needle's poles is reversed, causing the deflection of the needle to **reverse direction**. Correct option is (c).
PYQ 4Assertion-Reason — 1M | CBSE 2024; 2025
Topic 1
4. Assertion (A): Oersted's experiment established that electricity and magnetism are interrelated phenomena. Reason (R): A steady electric current in a conductor produces a magnetic field around it, which was first demonstrated by Oersted.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
Before Oersted's discovery, electricity and magnetism were studied as separate fields. Oersted's work linked them, starting electromagnetism. So Assertion (A) is true.
Step 2: Analyze Reason
Oersted showed that a current-carrying wire deflected a nearby compass, proving a steady current creates a magnetic field. So Reason (R) is true.
Step 3: Check Explanation
The fact that current (electricity) produces a magnetic field (magnetism) is the exact physical connection that established their interrelation. Correct option is (a).
PYQ 5SA — 2M | CBSE Recurring 2015–2024
Topic 1
5. Describe Oersted's experiment with a neat diagram. State the observation and conclusion. What does this experiment establish about the relationship between electricity and magnetism?
Final Answer: Current-carrying wire deflects compass needle, establishing that electric current creates magnetic fields.
Step 1: Setup
A magnetic compass needle is placed directly below or above a straight wire aligned in the North-South direction, connected to a battery and key.
Step 2: Observations
• When current flows, the compass needle deflects from its N-S alignment. • Reversing current reverses deflection. • Increasing current increases deflection.
Step 3: Conclusion
A current-carrying conductor generates a magnetic field in its surrounding space, establishing that moving electric charges are the source of magnetic fields.
PYQ 6SA — 2M | CBSE Recurring 2016–2025
Topic 1
6. Define magnetic field at a point. On what does the magnitude of the magnetic force on a moving charge depend? State the SI unit of magnetic field.
Final Answer: Region of influence around magnet/current. Force depends on q, v, B, angle. SI Unit is Tesla (T).
Step 1: Magnetic Field Definition
The magnetic field at a point is the space around a current-carrying conductor or magnet where its magnetic influence (force) can be experienced by moving charges or magnetic dipoles.
Step 2: Force Dependences
The magnetic Lorentz force is $F = q v B \sin\theta$. It depends on: 1. Magnitude of charge ($q$) 2. Speed of charge ($v$) 3. Strength of magnetic field ($B$) 4. Angle ($\theta$) between the velocity and field vector.
Step 3: SI Unit
SI Unit: **Tesla (T)** or $\text{N}\cdot\text{A}^{-1}\cdot\text{m}^{-1}$.
PYQ 7SA — 2M | CBSE 2017; 2020; Recurring
Topic 1
7. Write four similarities between Biot-Savart law and Coulomb's law.
Final Answer: Similarities include: long-range (1/r²), superpositions apply, field proportional to source strength.
Step 1: Inverse-square nature
Both fields (electric and magnetic) are **long-range fields** and obey the **inverse-square law** ($E \propto 1/r^2$ and $B \propto 1/r^2$).
Step 2: Superposition principle
Both laws obey the **principle of superposition**, where the net field is the vector sum of individual contributions.
Step 3: Source dependence
Both fields are **directly proportional** to the strength of their sources (charge $q$ in Coulomb's law, and current element $I\vec{dl}$ in Biot-Savart law).
Step 4: Wave propagation
Both fields propagate through space at the speed of light in vacuum.
TOPIC 2
Biot-Savart Law & Circular Loop
10 Questions
PYQ 8MCQ — 1M | CBSE Recurring 2017–2026
Topic 2
8. According to Biot-Savart law, the magnetic field dB due to a current element I dl at a distance r is proportional to:
(a) I dl sin θ / r²
(b) I dl sin θ / r
(c) I dl cos θ / r²
(d) I dl / r²
Correct Answer: (a) I dl sin θ / r²
Step 1: Write Biot-Savart Law
The magnitude of the magnetic field $dB$ due to current element $Id\vec{l}$ at distance $r$ is: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$$
Step 2: Identify proportionality
From the formula, $dB \propto \frac{I dl \sin\theta}{r^2}$. Correct option is (a).
PYQ 9MCQ — 1M | CBSE Recurring 2018–2026
Topic 2
9. The magnetic field at the centre of a circular current-carrying coil of radius R carrying current I (number of turns = N) is given by:
(a) B = μ₀NI/2R
(b) B = μ₀NI/4πR
(c) B = μ₀NI/R
(d) B = μ₀I/2πR
Correct Answer: (a) B = μ₀NI/2R
Step 1: Circular Loop formula
The magnetic field at the centre of a single-turn circular loop is $B = \frac{\mu_0 I}{2R}$.
Step 2: Extension to N turns
For a coil with $N$ closely wound turns, the magnetic fields due to individual turns add up constructively as they have the same direction: $$B = \frac{\mu_0 N I}{2R}$$ Correct option is (a).
PYQ 10MCQ — 1M | CBSE Board Recurring; 2019; 2022
Topic 2
10. Two identical circular coils of radius R carry currents of 1 A and √3 A and are placed concentrically, lying in the XY and YZ planes respectively. What is the magnitude of the net magnetic field at the centre of the coils?
(a) μ₀/R
(b) μ₀/2R
(c) √3 μ₀/2R
(d) 2 μ₀/R
Correct Answer: (a) μ₀/R
Step 1: Write B-field vectors
• Coil 1 in XY plane: Magnetic field is along Z-axis: $\vec{B}_1 = \frac{\mu_0 I_1}{2R} \hat{k} = \frac{\mu_0(1)}{2R} \hat{k}$ • Coil 2 in YZ plane: Magnetic field is along X-axis: $\vec{B}_2 = \frac{\mu_0 I_2}{2R} \hat{i} = \frac{\mu_0(\sqrt{3})}{2R} \hat{i}$
Step 2: Resultant Field
Since $\vec{B}_1$ and $\vec{B}_2$ are perpendicular: $$B_{\text{net}} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{2R} \sqrt{1^2 + (\sqrt{3})^2} = \frac{\mu_0}{2R} \sqrt{1 + 3} = \frac{\mu_0 \cdot 2}{2R} = \frac{\mu_0}{R}$$ Correct option is (a).
11. Assertion (A): The magnetic field at the centre of a circular current-carrying loop is perpendicular to the plane of the loop. Reason (R): Each current element produces a magnetic field at the centre whose direction, given by the right-hand rule, is perpendicular to the plane of the loop, and all such contributions add up.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
At the centre of a circular loop, the magnetic field lines form a straight line perpendicular to the plane of the loop. Assertion (A) is true.
Step 2: Analyze Reason
Using the right-hand rule, for any small element $d\vec{l}$ on the loop, the direction of $d\vec{l} \times \vec{r}$ points perpendicular to the plane of the loop. All these elements contribute in the same direction, so they sum up. Reason (R) is true.
Step 3: Check Explanation
Reason (R) explains the vector summation that gives the perpendicular net magnetic field. Thus, R is the correct explanation. Correct option is (a).
12. State Biot-Savart law. Write it in vector form. Name the physical quantity whose SI unit is tesla (T).
Final Answer: dB = (μ₀/4π) * (I dl × r) / r³. Tesla is unit of magnetic field.
Step 1: Statement
Biot-Savart law states that the magnetic field ($d\vec{B}$) produced by a small current element $Id\vec{l}$ at a point $P$ at distance $r$ is: 1. Directly proportional to current $I$. 2. Directly proportional to length $dl$. 3. Directly proportional to $\sin\theta$. 4. Inversely proportional to square of distance $r^2$.
The physical quantity is the **magnetic field** (or magnetic flux density, magnetic induction).
PYQ 13SA — 2M | CBSE 2016; 2019; Recurring
Topic 2
13. A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small current element $dl$ at point P situated at distance r from the element.
Final Answer: Perpendicular to both current element and position vector.
Step 1: Apply vector cross product
The direction of $d\vec{B}$ is given by the cross product $d\vec{l} \times \vec{r}$. Since $d\vec{l}$ is along the conductor (perpendicular to the paper) and $\vec{r}$ lies in the plane of the paper, their cross product is perpendicular to both.
Step 2: Direction identification
• If current is flowing outwards (pointing out of paper $\odot$), the field at point $P$ will be perpendicular to line $r$ pointing counter-clockwise (tangential). • If current flows inwards (pointing into paper $\otimes$), the field points clockwise (tangential).
PYQ 14LA — 5M | CBSE AI 2014; Comptt. All India 2017; Recurring 2016–2026
Topic 2
14. Using Biot-Savart law, derive an expression for the magnetic field at the centre of a circular coil of radius R, number of turns N, carrying current I. Draw the magnetic field lines due to the circular current loop.
Final Answer: Derivation showing B = μ₀NI/2R.
Step 1: Set up circular loop geometry
Consider a circular loop of radius $R$ carrying current $I$. Take a small element $d\vec{l}$ on the boundary. The distance from $d\vec{l}$ to the centre is $R$, and the angle $\theta$ between $d\vec{l}$ and position vector $\vec{R}$ is $90^\circ$ everywhere.
Step 2: Apply Biot-Savart Law
The magnitude of the magnetic field due to element $d\vec{l}$ at the centre is: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin 90^\circ}{R^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2}$$
Step 3: Integrate over loop
Since the field contributions due to all elements point in the same direction (perpendicular to loop plane), we integrate: $$B = \int dB = \frac{\mu_0 I}{4\pi R^2} \int dl$$ Since $\int dl = 2\pi R$ (circumference of loop): $$B = \frac{\mu_0 I}{4\pi R^2} (2\pi R) = \frac{\mu_0 I}{2R}$$
Step 4: Extension to N turns
For $N$ turns, the field is multiplied by $N$: $$B = \frac{\mu_0 N I}{2R}$$
PYQ 15LA — 5M | CBSE 2018; 2021; 2024; Recurring
Topic 2
15. (a) State Biot-Savart law in vector form. (b) Two identical circular coils A and B each of radius R carrying currents I and √3I are placed with their planes perpendicular to each other and with their centres coinciding. Find the magnitude and direction of the resultant magnetic field at the centre.
Final Answer: (b) B_net = μ₀I/R at 60° to XY plane.
Step 1: Part (a) Vector Form
Vector Biot-Savart law is: $$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3}$$
Step 2: Part (b) Field Components
• Field due to coil A (current $I$): $B_A = \frac{\mu_0 I}{2R}$ • Field due to coil B (current $\sqrt{3}I$): $B_B = \frac{\mu_0 \sqrt{3} I}{2R}$
Step 3: Resultant Field Magnitude
Since planes are perpendicular, the field vectors $\vec{B}_A$ and $\vec{B}_B$ are perpendicular: $$B_{\text{net}} = \sqrt{B_A^2 + B_B^2} = \frac{\mu_0 I}{2R} \sqrt{1^2 + (\sqrt{3})^2} = \frac{\mu_0 I}{2R} (2) = \frac{\mu_0 I}{R}$$
Step 4: Resultant Field Direction
Angle $\phi$ with respect to $\vec{B}_A$ is: $$\tan\phi = \frac{B_B}{B_A} = \frac{\sqrt{3}}{1} \implies \phi = 60^\circ$$
PYQ 16LA — 5M | CBSE 2019; 2022; 2025; Recurring
Topic 2
16. Derive an expression for the magnetic field on the axis of a circular current loop of radius R carrying current I at a distance x from the centre. What is the field at the centre? Draw a graph showing variation of B with x.
Final Answer: B = μ₀IR² / [2(R² + x²)^(3/2)]
Step 1: Axis Geometry Setup
Consider a loop of radius $R$ in the YZ plane carrying current $I$. Let $P$ be a point on the X-axis (axial point) at distance $x$ from the centre $O$. Let $r = \sqrt{R^2 + x^2}$ be the distance from a current element $d\vec{l}$ to point $P$.
Step 2: Resolve B-field components
The field $d\vec{B}$ due to element $d\vec{l}$ has magnitude $dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2}$ (since angle between $d\vec{l}$ and $\vec{r}$ is $90^\circ$). Resolve $d\vec{B}$: • Perpendicular to axis: $dB \sin\theta$ (cancels out by symmetry from diametrically opposite elements). • Parallel to axis: $dB \cos\theta$ (adds up constructively).
Step 3: Integrate along loop
Total field $B$ is: $$B = \int dB \cos\theta = \int \left( \frac{\mu_0}{4\pi} \frac{I dl}{r^2} \right) \frac{R}{r} = \frac{\mu_0 I R}{4\pi r^3} \int dl$$ $$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2\pi R) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
Step 4: Check at centre
At the centre, $x = 0$: $$B = \frac{\mu_0 I R^2}{2(R^2)^{3/2}} = \frac{\mu_0 I}{2R}$$ This matches the centre loop formula. The graph of $B$ vs $x$ is a symmetric bell-shaped curve peaking at $x=0$.
PYQ 17Case Study — 4M | CBSE 2023; 2024; 2025
Topic 2
17. Case Study — Biot-Savart Law and Circular Loop: Biot-Savart law states that the magnetic field dB⃗ due to a small current element I dl⃗ at a distance r is: dB = (μ₀/4π)(I dl sin θ/r²). For a circular loop of radius R carrying current I, every element contributes a dB perpendicular to the plane of the loop. By symmetry, all horizontal components cancel, and the vertical components add up to give B = μ₀I/2R at the centre. For N turns, B = μ₀NI/2R. (i) State Biot-Savart law. Write its SI unit. (ii) A circular coil of 50 turns, radius 0.2 m carries 5 A. Find the magnetic field at its centre. (iii) How does the magnetic field at the centre change if the number of turns is doubled but the current is halved? (iv) Draw the magnetic field pattern due to a single-turn current-carrying circular loop.
Final Answer: (i) Tesla (T), (ii) 7.85 × 10⁻⁴ T, (iii) Unchanged, (iv) Closed loops.
Step 1: Part (i)
Biot-Savart Law expression: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \hat{r})}{r^2}$. SI Unit: Tesla (T).
Since $B \propto N I$. If $N' = 2N$ and $I' = I/2$: $$B' \propto (2N)(I/2) = NI \implies B' = B$$ The magnetic field remains **unchanged**.
Step 4: Part (iv)
The magnetic field lines form concentric circular loops around every segment of the wire, passing through the inside of the loop in one direction and looping back around the outside.
PYQ 92LA — 5M | CBSE 2016; 2019; 2022; Recurring
Topic 2
92. (a) State Biot-Savart law. Using it, derive an expression for the magnetic field at the centre of a circular current loop of radius R and N turns carrying current I. (b) Draw the field lines. (c) Two coils A and B, each of 100 turns and radius 20 cm, are placed concentrically, with their planes perpendicular. They carry currents of 1 A and √3 A respectively. Find the net magnetic field at the centre.
Final Answer: (c) $2\pi \times 10^{-4}\text{ T}$
Step 1: Part (a) & (b) Derivations & Diagrams
• Derivation: See Q14. • Field lines: Represented by closed circular loops wrapping around the conductor wire.
Since the planes are perpendicular, the field vectors are orthogonal: $$B_{\text{net}} = \sqrt{B_A^2 + B_B^2} = \pi \times 10^{-4} \sqrt{1^2 + (\sqrt{3})^2} = 2\pi \times 10^{-4}\text{ T}$$
PYQ 99LA — 5M | CBSE 2018; 2022; 2025; Recurring
Topic 2
99. (a) State Biot-Savart law. (b) Using it, derive the magnetic field at any point on the axis of a circular loop of radius R carrying current I. (c) Show that at large distances (x >> R), the field resembles that of a magnetic dipole. Find the expression for the magnetic dipole moment.
Final Answer: (c) B = μ₀m / (2πx³)
Step 1: Part (a) & (b) Derivation
Refer to Q12 and Q16 for Biot-Savart law and axial circular loop derivation.
Step 2: Part (c) Axial field approximation for x >> R
If $x \gg R$, then $R^2 + x^2 \approx x^2$. The axial field formula simplifies to: $$B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \approx \frac{\mu_0 I R^2}{2 x^3}$$
Step 3: Link to magnetic moment
Multiply numerator and denominator by $2\pi$: $$B = \frac{\mu_0}{4\pi} \frac{2 I (\pi R^2)}{x^3} = \frac{\mu_0}{4\pi} \frac{2 m}{x^3}$$ where $m = I A = I(\pi R^2)$ is the magnetic dipole moment of the loop. This has the identical mathematical form as the electric field on the axial line of an electric dipole ($E = \frac{1}{4\pi\varepsilon_0} \frac{2p}{x^3}$).
PYQ 102LA — 5M | CBSE 2020; 2023; 2025; Recurring
Topic 2
102. (a) Derive the expression for the magnetic field on the axis of a current-carrying circular loop. (b) At the centre (x = 0), write the field formula. (c) At x >> R, show it resembles a dipole field. (d) A circular loop of radius 8 cm carries a current of $\pi\text{ A}$. Find the magnetic field at: (i) centre, (ii) a point 6 cm along the axis from the centre.
Final Answer: (d) (i) $2.46 \times 10^{-5}\text{ T}$, (ii) $1.26 \times 10^{-5}\text{ T}$
Step 1: Part (a), (b), (c) Explanations
Refer to Q16 and Q99 for the derivations and limits.
18. Ampere's circuital law states that the line integral of the magnetic field B⃗ around a closed loop is equal to:
(a) μ₀ times the total current enclosed by the loop
(b) ε₀ times the total current enclosed
(c) μ₀/ε₀ times the total charge enclosed
(d) μ₀I² for the loop
Correct Answer: (a) μ₀ times the total current enclosed by the loop
Step 1: Ampere's Law Statement
Ampere's circuital law states that the line integral of magnetic field $\vec{B}$ around any closed path (Amperian loop) is equal to $\mu_0$ times the total current $I_{\text{encl}}$ passing through the surface bounded by the closed path: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{encl}}$$ Correct option is (a).
PYQ 19MCQ — 1M | CBSE Recurring 2016–2026
Topic 3
19. The magnetic field at a perpendicular distance r from an infinitely long straight wire carrying current I is:
(a) B = μ₀I/4πr
(b) B = μ₀I/2πr
(c) B = μ₀I/2r
(d) B = 2μ₀I/πr
Correct Answer: (b) B = μ₀I/2πr
Step 1: Straight Wire Field formula
The magnetic field at distance $r$ due to a long straight wire carrying current $I$ is: $$B = \frac{\mu_0 I}{2\pi r}$$ Correct option is (b).
PYQ 20MCQ — 1M | CBSE/NCERT Recurring
Topic 3
20. A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field at a point 20 cm from the wire?
21. Assertion (A): Ampere's circuital law is not independent of the Biot-Savart law. Reason (R): Ampere's circuital law can be derived from the Biot-Savart law. Its relationship to Biot-Savart law is similar to that between Gauss's law and Coulomb's law.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
Ampere's law and Biot-Savart law are both descriptions of the same magnetic phenomenon and are consistent. Assertion (A) is true.
Step 2: Analyze Reason
Just as Gauss's law is an integral form of electrostatics derived from Coulomb's law, Ampere's law is an integral formulation derived from Biot-Savart law. Reason (R) is true.
Step 3: Check Explanation
Since R establishes that Ampere's law can be derived from Biot-Savart law, it shows why they are not independent. Correct option is (a).
22. Assertion (A): Ampere's circuital law holds good only for steady currents. Reason (R): For a changing current, the displacement current term must be added (Maxwell's modification), and the law in its original form is not valid.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
The original form of Ampere's law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$ leads to contradictions when applied to time-varying electric fields (e.g. charging capacitor). So it is valid only for steady currents. Assertion (A) is true.
Step 2: Analyze Reason
Maxwell modified the law to $\oint \vec{B}\cdot d\vec{l} = \mu_0(I_c + I_d)$, where $I_d = \varepsilon_0 \frac{d\Phi_e}{dt}$ is the displacement current. Without this, the original law is incomplete. Reason (R) is true.
Step 3: Check Explanation
Reason (R) states the mathematical correction required for changing fields, which directly explains the limitation of the original law. Correct option is (a).
PYQ 23SA — 2M | CBSE Recurring 2015–2026
Topic 3
23. State Ampere's circuital law. Write its mathematical form. State one limitation of this law.
Final Answer: ∮ B · dl = μ₀I_encl. Limitation: Not valid for changing fields.
Step 1: Statement & Equation
The line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_0$ times the net current enclosed by that path: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{encl}}$$
Step 2: Limitations
The original law assumes steady currents and **does not hold for non-steady currents** or circuits containing elements with time-varying electric fields (like charging capacitors) where displacement current is present.
PYQ 24LA — 5M | CBSE Recurring 2015–2025; 2026
Topic 3
24. Using Ampere's circuital law, derive an expression for the magnetic field at a perpendicular distance r from an infinitely long straight wire carrying current I. Draw the necessary diagram and show the direction of the field.
Final Answer: Derivation of B = μ₀I/2πr.
Step 1: Set up Amperian loop
Consider an infinitely long straight wire carrying current $I$. To find the field at distance $r$, draw a circular Amperian loop of radius $r$ concentric with the wire. The magnetic field $\vec{B}$ is tangential to the circle at every point, meaning $\theta = 0^\circ$ between $\vec{B}$ and $d\vec{l}$.
Step 2: Apply Ampere's Law
$$\oint \vec{B} \cdot d\vec{l} = \oint B dl \cos 0^\circ = B \oint dl = B(2\pi r)$$
Step 3: Equate and Solve
According to Ampere's law, this integral is equal to $\mu_0 I$: $$B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$$
Step 4: Direction
The direction of the field lines is circular, determined by the **right-hand thumb rule** (thumb along current, curled fingers along field lines).
PYQ 25LA — 5M | CBSE 2017; 2020; 2023; Recurring
Topic 3
25. (a) State Ampere's circuital law. (b) Apply it to find the magnetic field at a distance r from the axis of a long cylindrical conductor of radius a carrying uniformly distributed current I for: (i) r > a (outside), and (ii) r < a (inside). Plot a graph of B versus r.
Final Answer: (i) B_out = μ₀I/2πr, (ii) B_in = μ₀Ir/2πa².
Step 1: Outside cylinder (r > a)
Draw an Amperian loop of radius $r > a$. The entire current $I$ is enclosed: $$\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = \mu_0 I \implies B_{\text{out}} = \frac{\mu_0 I}{2\pi r} \quad \left(B \propto \frac{1}{r}\right)$$
Step 2: Inside cylinder (r < a)
Draw an Amperian loop of radius $r < a$. The current enclosed is proportional to the area: $$I_{\text{encl}} = I \left( \frac{\pi r^2}{\pi a^2} \right) = I \frac{r^2}{a^2}$$ Apply Ampere's law: $$B(2\pi r) = \mu_0 I_{\text{encl}} = \mu_0 I \frac{r^2}{a^2} \implies B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2} \quad (B \propto r)$$
Step 3: Graph B vs r
• Inside ($r < a$): $B$ increases linearly from $0$ at the axis to $\frac{\mu_0 I}{2\pi a}$ at the surface. • Outside ($r > a$): $B$ decreases hyperbolically. Peak value occurs at the surface $r = a$.
PYQ 26SA — 3M | CBSE 2019; 2022; Recurring
Topic 3
26. A long straight wire of circular cross-section of radius a carries a steady current I distributed uniformly across its cross-section. Using Ampere's law, find the magnetic field: (i) at a distance r > a from the axis, (ii) at the surface (r = a), and (iii) at a distance r < a from the axis.
Final Answer: (i) B_out = μ₀I/2πr, (ii) B_surface = μ₀I/2πa, (iii) B_in = μ₀Ir/2πa².
Step 1: Outside (r > a)
As derived in Q25, $B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$.
Step 2: Surface (r = a)
Substituting $r = a$ in either formula gives: $$B_{\text{surface}} = \frac{\mu_0 I}{2\pi a}$$
Step 3: Inside (r < a)
As derived in Q25, $B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2}$.
27. Case Study — Ampere's Circuital Law: Ampere's circuital law states: ∮ B⃗ · dl⃗ = μ₀I_enclosed. It is the magnetic analog of Gauss's law in electrostatics. For an infinitely long straight wire carrying current I, a circular Amperian loop of radius r gives: B(2πr) = μ₀I, so B = μ₀I/2πr. The field lines are concentric circles around the wire. The direction is given by the right-hand thumb rule. (i) State Ampere's circuital law in words and in mathematical form. (ii) Find the magnetic field at a distance 10 cm from a long wire carrying 5 A current. (iii) Draw the magnetic field pattern around a long straight current-carrying wire. (iv) On what principle is Ampere's law based? What conservation law does it support?
Final Answer: (i) ∮ B · dl = μ₀I, (ii) 1.0 × 10⁻⁵ T, (iii) Circular lines, (iv) Magnetic field lines close.
Step 1: Part (i)
Ampere's law states that the line integral of magnetic field around a closed loop equals $\mu_0$ times the enclosed current: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I$$
Concentric circles centered on the wire, lying in the plane perpendicular to the wire. Direction is given by the right-hand thumb rule.
Step 4: Part (iv)
It is based on the idea that magnetic field lines are continuous closed loops (non-existence of isolated magnetic monopoles). It supports conservation of current (charge) flow in steady state.
PYQ 93LA — 5M | CBSE 2017; 2021; 2024; Recurring
Topic 3
93. (a) Using Ampere's circuital law, derive an expression for the magnetic field due to an infinitely long straight wire carrying current I at a distance r from it. (b) Find the magnetic field at a point 10 cm from a long wire carrying 15 A. (c) If two such parallel wires are 20 cm apart and carry 15 A each in the same direction, find the force per unit length between them.
Step 3: Part (c) Force per unit length calculation
Given $I_1 = I_2 = 15\text{ A}$, $d = 20\text{ cm} = 0.2\text{ m}$: $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2 \times 10^{-7} \times 15 \times 15}{0.2} = 2.25 \times 10^{-4}\text{ N/m}$$ Since currents are in the same direction, the force is **attractive**.
TOPIC 4
Straight Solenoid (Qualitative)
6 Questions
PYQ 28MCQ — 1M | CBSE Recurring 2017–2026
Topic 4
28. The magnetic field inside a long straight solenoid carrying current I is:
(a) Zero at the centre
(b) Non-uniform and directed radially
(c) Uniform and directed along the axis
(d) Maximum at the ends and zero at the centre
Correct Answer: (c) Uniform and directed along the axis
Step 1: Field inside Solenoid
A long solenoid consists of closely wound coils of wire. Inside the solenoid, the magnetic fields from individual turns reinforce each other along the axis, and cancel each other out in the radial direction.
Step 2: Conclusion
This results in a strong, **uniform magnetic field directed along the axis** of the solenoid. Correct option is (c).
PYQ 29MCQ — 1M | CBSE Recurring
Topic 4
29. The magnetic field inside an ideal solenoid (n turns per metre, current I) is given by:
(a) B = μ₀nI
(b) B = μ₀n²I
(c) B = μ₀nI/2
(d) B = μ₀I/2n
Correct Answer: (a) B = μ₀nI
Step 1: Solenoid field equation
For a long ideal solenoid, the magnetic field inside is constant and given by: $$B = \mu_0 n I$$ where $n = N/L$ is the number of turns per unit length. Correct option is (a).
PYQ 30Assertion-Reason — 1M | CBSE 2024; 2025
Topic 4
30. Assertion (A): The magnetic field outside a long ideal solenoid is nearly zero. Reason (R): In an ideal solenoid, all the magnetic field lines are confined inside the solenoid and there is nearly no field outside, similar to a bar magnet with north and south poles at the two ends.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
For an infinitely long, closely wound solenoid, the magnetic field outside is negligible (approaching zero). So Assertion (A) is true.
Step 2: Analyze Reason
The geometry causes outside magnetic field loops to spread out over infinite space, reducing their density (strength) to zero. The field lines are confined to the interior. So Reason (R) is true.
Step 3: Check Explanation
The confinement of field lines inside explains why the field outside is zero. Correct option is (a).
31. Draw the magnetic field lines due to a current flowing through a long solenoid. In what way does a solenoid behave like a bar magnet? Identify the north and south poles.
Final Answer: One end acts as North pole and the other as South pole, identical to a bar magnet.
Step 1: Bar Magnet Analogy
The magnetic field pattern of a current-carrying solenoid is identical to that of a bar magnet. It has magnetic field lines emerging from one end (North pole) and entering at the other end (South pole), forming closed loops.
Step 2: Pole Identification
Use the **clock rule** on the circular faces of the solenoid: • If current direction at a face is **clockwise**, that face behaves as a **South (S) pole**. • If **counter-clockwise**, it behaves as a **North (N) pole**.
PYQ 32SA — 2M | CBSE Recurring 2015–2025
Topic 4
32. What is a solenoid? State the two main properties of the magnetic field inside a long straight solenoid. Write the formula for magnetic field inside an ideal solenoid.
Final Answer: Helical coil of wire. Field is uniform and axial inside. B = μ₀nI.
Step 1: Definition
A solenoid is a long coil of wire wound in a closely spaced helix around a cylindrical frame, designed to produce a strong, controlled magnetic field inside when an electric current passes through it.
Step 2: Key Properties of Field Inside
1. Uniformity: The field is constant in magnitude and direction at all points inside the central region. 2. Direction: The field lines are parallel straight lines along the axis of the solenoid.
Step 3: Formula
$$B = \mu_0 n I$$
PYQ 33SA — 3M | CBSE 2016; 2019; 2022; Recurring
Topic 4
33. Using Ampere's circuital law, deduce an expression for the magnetic field inside a long solenoid having n turns per unit length and carrying current I. Draw the necessary diagram.
Final Answer: Derivation showing B = μ₀nI using rectangular loop.
Step 1: Set up rectangular Amperian loop
Consider a long solenoid carrying current $I$. Draw a rectangular Amperian loop $PQRS$ of length $L$ such that side $PQ$ lies inside the solenoid, parallel to the axis, and side $RS$ lies outside where $B=0$. Sides $QR$ and $SP$ are perpendicular to the axis.
Step 2: Evaluate line integral
$$\oint \vec{B} \cdot d\vec{l} = \int_P^Q \vec{B} \cdot d\vec{l} + \int_Q^R \vec{B} \cdot d\vec{l} + \int_R^S \vec{B} \cdot d\vec{l} + \int_S^P \vec{B} \cdot d\vec{l}$$ • $\int_P^Q B dl \cos 0^\circ = B L$ • $\int_Q^R B dl \cos 90^\circ = 0$ • $\int_R^S (0) dl = 0$ (outside field is zero) • $\int_S^P B dl \cos 90^\circ = 0$ Hence, $\oint \vec{B} \cdot d\vec{l} = B L$.
Step 3: Apply Ampere's Law
The number of turns enclosed by the loop is $n L$, so the total current enclosed is $I_{\text{encl}} = n L I$. $$B L = \mu_0 (n L I) \implies B = \mu_0 n I$$
PYQ 96LA — 5M | CBSE 2016; 2020; 2023; Recurring
Topic 4
96. (a) State Ampere's circuital law. Deduce the expression for the magnetic field inside a long solenoid having n turns per metre carrying current I. (b) Draw the field lines for the solenoid. (c) Two coaxial solenoids with 500 turns each, radii 3 cm and 5 cm, carry currents of 2 A and 3 A in opposite directions. Find the net magnetic field inside both, at a point on the common axis.
Final Answer: (c) $2.51 \times 10^{-3}\text{ T}$
Step 1: Part (a) & (b) Explanations
Refer to Q23, Q31 and Q33 for Ampere's law, solenoid derivation, and field lines diagram.
Step 2: Part (c) Solenoid specifications
Let length of solenoids be $L$ (assume long solenoids so $n = N/L$). If lengths are not specified, let's write the formula in terms of $n$. Let's assume turns per unit length $n = 500\text{ turns/m}$ for both. • $B_1 = \mu_0 n_1 I_1 = (4\pi \times 10^{-7}) \times 500 \times 2 = 4\pi \times 10^{-4}\text{ T} \approx 1.26 \times 10^{-3}\text{ T}$ • $B_2 = \mu_0 n_2 I_2 = (4\pi \times 10^{-7}) \times 500 \times 3 = 6\pi \times 10^{-4}\text{ T} \approx 1.88 \times 10^{-3}\text{ T}$
Step 3: Net magnetic field
Since the currents flow in opposite directions, the axial magnetic fields oppose each other: $$B_{\text{net}} = |B_2 - B_1| = 2\pi \times 10^{-4}\text{ T} \approx 6.28 \times 10^{-4}\text{ T}$$ If length $L = 1\text{ m}$, this is correct. If the question implies $N=500$ turns over a specific length $L = 0.5\text{ m}$, then $n = 1000\text{ turns/m}$, and the field is doubled.
103. (a) State Ampere's law. (b) Apply it to find B inside a solenoid of n turns/m, current I. (c) A solenoid 0.5 m long has 500 turns and carries 2 A. Find the magnetic field inside. (d) If a material of relative permeability μᵣ = 1000 is inserted, what is the new field? Explain the role of μᵣ.
Final Answer: (c) $2.51 \times 10^{-3}\text{ T}$, (d) $2.51\text{ T}$
Step 1: Part (a) & (b) Explanations
Refer to Q23 and Q33 for statements and derivation.
Step 2: Part (c) Solenoid Calculation
Given $L = 0.5\text{ m}$, $N = 500$, $I = 2\text{ A}$. Turns per unit length: $n = N/L = 500 / 0.5 = 1000\text{ turns/m}$. $$B_0 = \mu_0 n I = (4\pi \times 10^{-7}) \times 1000 \times 2 = 8\pi \times 10^{-4}\text{ T} \approx 2.51 \times 10^{-3}\text{ T}$$
Step 3: Part (d) Medium insertion
With a magnetic core of relative permeability $\mu_r = 1000$, the field is multiplied by $\mu_r$: $$B = \mu_r B_0 = 1000 \times 2.51 \times 10^{-3}\text{ T} = 2.51\text{ Tesla}$$ The relative permeability $\mu_r$ acts to align the microscopic magnetic domains of the material with the solenoid's field, dramatically amplifying the net magnetic flux density.
TOPIC 5
Force on Moving Charge (Lorentz)
12 Questions
PYQ 34MCQ — 1M | CBSE Recurring 2016–2026
Topic 5
34. The force on a charge q moving with velocity v in a magnetic field B is given by:
(a) F = qv + B
(b) F = q(v × B)
(c) F = q(v · B)
(d) F = q(B × v)/v
Correct Answer: (b) F = q(v × B)
Step 1: Magnetic Lorentz Force
The force experienced by a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the cross product: $$\vec{F} = q(\vec{v} \times \vec{B})$$ Correct option is (b).
PYQ 35MCQ — 1M | CBSE Recurring 2018–2026
Topic 5
35. A charged particle moves in a magnetic field. The work done by the magnetic force on the particle is:
Since magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$ (due to cross product definition), we have $\vec{F} \cdot \vec{v} = 0$. Thus, the rate of work done (power) is zero, so the net work done is **zero**. Correct option is (c).
PYQ 36MCQ — 1M | CBSE Recurring 2019–2026
Topic 5
36. A proton moving with velocity v enters a uniform magnetic field B directed into the page. The proton is moving towards the right (along +x axis). The magnetic force on the proton is directed:
(a) Along +x axis
(b) Along −x axis
(c) Along +y axis
(d) Along −y axis
Correct Answer: (c) Along +y axis
Step 1: Vector directions
• Velocity $\vec{v} = v \hat{i}$ (right) • Magnetic field $\vec{B} = -B \hat{k}$ (into the page) • Proton charge is positive ($q > 0$).
Step 2: Compute cross product
$$\vec{F} = q(\vec{v} \times \vec{B}) = q(v \hat{i} \times (-B \hat{k})) = -q v B (\hat{i} \times \hat{k})$$ Since $\hat{i} \times \hat{k} = -\hat{j}$: $$\vec{F} = -q v B (-\hat{j}) = q v B \hat{j}$$ This points in the $+y$ direction (upward). Correct option is (c).
PYQ 37MCQ — 1M | CBSE Recurring 2018–2025
Topic 5
37. A charged particle moving perpendicular to a uniform magnetic field describes a circular path. If the speed of the particle is doubled (keeping B and charge the same), the radius of the circular path:
(a) Remains unchanged
(b) Doubles
(c) Halves
(d) Quadruples
Correct Answer: (b) Doubles
Step 1: Write circular orbit radius formula
For centripetal force provided by magnetic force: $$\frac{m v^2}{r} = q v B \implies r = \frac{m v}{q B}$$
Step 2: Proportionality scaling
Since $m, q, B$ are constant, $r \propto v$. Doubling speed $v$ will **double** the radius $r$. Correct option is (b).
PYQ 38MCQ — 1M | CBSE Board Recurring
Topic 5
38. A proton and an electron travel along parallel paths and enter a region of uniform magnetic field B acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?
(a) Proton
(b) Electron
(c) Both have equal frequency
(d) Neither, they travel in straight lines
Correct Answer: (b) Electron
Step 1: Write frequency formula
The frequency of revolution in a magnetic field (cyclotron frequency) is: $$f = \frac{q B}{2\pi m}$$
Step 2: Compare parameters
Both electron and proton have equal charge magnitude ($q = 1.6 \times 10^{-19}\text{ C}$). Since $f \propto 1/m$, the lighter particle will have a higher frequency. The mass of an electron is much smaller than a proton ($m_e \approx m_p/1836$). Thus, the **electron** will revolve with a higher frequency. Correct option is (b).
39. Assertion (A): A magnetic field exerts no force on a stationary electric charge. Reason (R): The magnetic force on a charge is F = q(v × B), which is zero when v = 0.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
Magnetic fields only interact with moving charges. A charge at rest ($v=0$) feels no force. So Assertion (A) is true.
Step 2: Analyze Reason
The Lorentz force equation is $F = q v B \sin\theta$. If $v = 0$, $F = 0$. So Reason (R) is true.
Step 3: Check Explanation
The zero velocity is the mathematical reason why force is zero. So R explains A. Correct option is (a).
40. Assertion (A): A magnetic force cannot do work on a moving charge. Reason (R): The magnetic force on a moving charge is always perpendicular to its velocity, so the component of force in the direction of motion is zero; hence work done is zero.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
Since magnetic force does not change the kinetic energy/speed of a charge, the work done is always zero. Assertion (A) is true.
Step 2: Analyze Reason
Work is $W = \vec{F} \cdot d\vec{s}$. Since $\vec{F} \perp \vec{v}$, $\vec{F} \cdot d\vec{s} = 0$. Reason (R) is true.
Step 3: Check Explanation
The perpendicular orientation is the direct physical explanation for zero work done. Correct option is (a).
PYQ 41SA — 2M | CBSE 2018; 2020; 2022; Recurring
Topic 5
41. Write the relation for the force acting on a charge carrier q moving with velocity $\vec{v}$ through a magnetic field $\vec{B}$ in vector notation. Using this relation, deduce the conditions under which this force will be: (i) maximum, and (ii) minimum (zero).
Final Answer: F = q(v × B). Max when θ = 90°, min when θ = 0° or 180°.
Step 1: Vector Formula
$$\vec{F} = q(\vec{v} \times \vec{B}) \implies F = q v B \sin\theta$$ where $\theta$ is the angle between velocity and magnetic field vectors.
Step 2: Condition for Maximum Force
Force is maximum when $\sin\theta = 1 \implies \theta = 90^\circ$ (charge moves perpendicular to the field): $$F_{\text{max}} = q v B$$
Step 3: Condition for Minimum Force (Zero)
Force is zero when $\sin\theta = 0 \implies \theta = 0^\circ$ or $180^\circ$ (charge moves parallel or anti-parallel to the field): $$F_{\text{min}} = 0$$
42. A proton with kinetic energy $1.3384 \times 10^{-14}\text{ J}$ moving horizontally from north to south enters a uniform magnetic field B = 2.0 mT directed eastward. Calculate the force on the proton and describe its subsequent motion.
Final Answer: $1.28 \times 10^{-13}\text{ N}$ upwards.
Since velocity is North-to-South and field is Eastward, they are perpendicular ($\theta = 90^\circ$): $$F = q v B = (1.6 \times 10^{-19}) \times (4 \times 10^6) \times (2.0 \times 10^{-3}) = 1.28 \times 10^{-13}\text{ N}$$
Step 3: Determine direction and subsequent motion
Using Fleming's Left-hand Rule: • Field: East (index finger) • Motion/Current: South (middle finger) • Force points **vertically upwards** (thumb). Since force is perpendicular to velocity, the proton will move in a **vertical circular path**.
PYQ 43SA — 3M | CBSE Board recurring
Topic 5
43. Find the condition under which charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select particles of a particular speed. Draw a labelled diagram of such a device (velocity selector).
Final Answer: v = E/B when electric and magnetic forces balance.
Step 1: Crossed Fields principle
Apply electric field $\vec{E}$ and magnetic field $\vec{B}$ perpendicular to each other and to the direction of motion of the incoming particles (crossed fields configuration).
Step 2: Balance of forces
The electric force $F_e = qE$ and magnetic force $F_m = qvB$ act in opposite directions. For a particle to pass undeflected, these forces must balance: $$q E = q v B \implies v = \frac{E}{B}$$
Step 3: Conclusion
Only particles with this exact speed $v = E/B$ will experience zero net force and pass through the exit slit undeflected. Others will be deflected and filtered out.
44. A charged particle enters a region where electric field $\vec{E}$ and magnetic field $\vec{B}$ are perpendicular to each other and to the direction of motion of the particle. Show that the particle will pass through the region undeflected if v = E/B. For what class of particles is this a velocity selector?
Final Answer: v = E/B. Works for all charged particles regardless of mass/charge.
Step 1: Net force equation
The total Lorentz force is $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. Since $\vec{E}$ and $\vec{v} \times \vec{B}$ are collinear but opposite: $$F_{\text{net}} = q E - q v B$$
Step 2: Undeflected condition
For $F_{\text{net}} = 0$, we have $q E = q v B \implies v = E/B$.
Step 3: Independence of Charge/Mass
Since $q$ cancels out, this condition is independent of the charge, mass, and sign of the particle. It acts as a velocity selector for **any charged particles** (electrons, protons, alpha particles, etc.).
45. Case Study — Lorentz Force on a Moving Charge: When a charge q moves with velocity v in a region of electric field E and magnetic field B, the total force experienced is the Lorentz force: F = q(E + v × B). The magnetic component qv × B is always perpendicular to v, so it does no work on the charge and only changes the direction of motion. If the particle moves perpendicular to B, it follows a circular path of radius r = mv/qB. The angular frequency (cyclotron frequency) is ω = qB/m, which is independent of speed. (i) Write the expression for Lorentz force on a charge moving in combined E and B fields. (ii) Under what condition does a magnetic force perform no work on a moving charge? (iii) A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves perpendicular to B = 0.5 T with speed 10⁶ m/s. Find the radius of its circular path. (iv) A velocity selector has E = 10⁴ V/m and B = 0.1 T. What is the speed of the selected particle?
Final Answer: (i) F = q(E + v × B), (ii) Always, (iii) 2.09 cm, (iv) 10⁵ m/s
Step 1: Part (i)
$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$
Step 2: Part (ii)
A magnetic force always performs zero work on a charge because the force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to the instantaneous velocity $\vec{v}$ (meaning $\vec{F} \cdot d\vec{r} = 0$).
98. (a) Write the Lorentz force equation. A charged particle enters a region of uniform magnetic field B perpendicular to its motion. Derive the expression for radius r = mv/qB of its circular orbit. (b) Compare for proton and deuteron: (i) radii, (ii) time periods, (iii) frequencies, if they enter with same speed. (c) A proton moves in B = 0.3 T with KE = $8 \times 10^{-15}\text{ J}$. Find radius.
101. (a) What is a velocity selector? With a neat diagram, explain how crossed electric and magnetic fields are used to select charged particles of a specific speed. (b) A velocity selector uses E = $2 \times 10^4\text{ V/m}$ and B = 0.4 T. Find the speed of the selected particles. (c) If the electric field is switched off, describe the subsequent motion of the particles in just B.
Final Answer: (b) $5 \times 10^4\text{ m/s}$, (c) Circular path.
If $\vec{E}$ is switched off, only the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts. Since velocity is perpendicular to the magnetic field, the force acts as a centripetal force, causing the particles to travel in a **uniform circular path** of radius $r = \frac{mv}{qB}$ in the plane perpendicular to $\vec{B}$.
TOPIC 6
Force on Current-Carrying Conductor
7 Questions
PYQ 46MCQ — 1M | CBSE Recurring 2016–2026
Topic 6
46. The force on a current-carrying conductor of length L, carrying current I, placed in a uniform magnetic field B at angle θ to the field, is:
(a) F = BIL cos θ
(b) F = BIL sin θ
(c) F = BIL/sin θ
(d) F = B²IL
Correct Answer: (b) F = BIL sin θ
Step 1: Conductor Force equation
The magnetic force on a straight current-carrying wire of length $\vec{L}$ is: $$\vec{F} = I(\vec{L} \times \vec{B})$$
Step 2: Magnitude
The magnitude of the cross product is: $$F = B I L \sin\theta$$ Correct option is (b).
PYQ 47MCQ — 1M | CBSE Recurring
Topic 6
47. A straight conductor carrying current I is placed perpendicular to a uniform magnetic field B. The force on the conductor is:
(a) Zero
(b) Maximum
(c) Directed along the current
(d) Directed along the field
Correct Answer: (b) Maximum
Step 1: Apply force equation
Since the conductor is perpendicular, $\theta = 90^\circ$ $\implies \sin 90^\circ = 1$.
Step 2: Conclusion
The force magnitude $F = B I L$ is at its **maximum value**. The direction is perpendicular to both current and field (given by Fleming's left-hand rule). Correct option is (b).
PYQ 48SA — 2M | CBSE Board Recurring 2017; 2020
Topic 6
48. Two wires of equal lengths are bent in the form of two loops — one square and one circular. Both carry the same current and are placed in the same uniform magnetic field. Which loop will experience a greater torque? Why?
(a) Circular loop
(b) Square loop
(c) Both equal
(d) Neither feels torque
Final Answer: (a) Circular loop
Step 1: Torque formula
Torque is given by $\tau = N I A B \sin\theta$. For identical turns, current, and field, torque is directly proportional to loop area $A$ ($\tau \propto A$).
Since $A_c > A_s$, the **circular loop** experiences a greater torque. Correct option is (a).
PYQ 49Assertion-Reason — 1M | CBSE 2023; 2025
Topic 6
49. Assertion (A): A current-carrying conductor placed parallel to a uniform magnetic field experiences no force. Reason (R): The magnetic force on a current-carrying conductor is F = BIL sin θ. When the conductor is parallel to the field, θ = 0°, so F = 0.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
If current flows along the direction of field lines, the magnetic force is zero. Assertion (A) is true.
Step 2: Analyze Reason
The force is given by $F = B I L \sin\theta$. When parallel, $\theta = 0^\circ \implies \sin 0^\circ = 0 \implies F = 0$. Reason (R) is true.
Step 3: Check Explanation
The mathematical reason explains why the force disappears when parallel. Correct option is (a).
PYQ 50SA — 3M | CBSE Recurring 2015–2025
Topic 6
50. Derive the expression for the force on a straight current-carrying conductor of length L, carrying current I, placed in a uniform magnetic field B. Write the vector form. Under what conditions is the force (i) maximum and (ii) zero?
Final Answer: F = I(L × B). Max when θ = 90°, zero when θ = 0°.
Step 1: Relate to microscopic forces
Let the conductor have cross-sectional area $A$ and carrier density $n$. The number of free electrons in length $L$ is $N = n A L$. The magnetic force on a single electron moving with drift velocity $\vec{v}_d$ is: $$\vec{f} = -e (\vec{v}_d \times \vec{B})$$
Step 2: Compute net force
The total force is $\vec{F} = N \vec{f} = (n A L) [-e (\vec{v}_d \times \vec{B})]$. Since current density vector is $\vec{I} = -n e A \vec{v}_d$, we can define length vector $\vec{L}$ in the direction of current: $$\vec{F} = I (\vec{L} \times \vec{B})$$
Step 3: Limits
• Maximum: $\theta = 90^\circ \implies F = B I L$ (conductor perpendicular to field). • Zero: $\theta = 0^\circ$ or $180^\circ \implies F = 0$ (conductor parallel to field).
51. A circular coil with cross-sectional area $0.2\text{ cm}^2$, carrying a current of $4\text{ A}$, is kept in a uniform magnetic field of magnitude 0.5 T normal to the plane of the coil. Calculate: (i) the maximum torque on the coil, (ii) the net force on the coil, and (iii) the torque when the plane of the coil makes 30° with the field.
Final Answer: (i) $4 \times 10^{-5}\text{ N·m}$, (ii) $0\text{ N}$, (iii) $3.46 \times 10^{-5}\text{ N·m}$
Step 1: Parameters Setup
Given $A = 0.2\text{ cm}^2 = 0.2 \times 10^{-4}\text{ m}^2 = 2 \times 10^{-5}\text{ m}^2$, $I = 4\text{ A}$, $B = 0.5\text{ T}$. Magnetic moment $m = I A = 4 \times 2 \times 10^{-5} = 8 \times 10^{-5}\text{ A\cdot m}^2$.
Step 2: Torque and Force calculations
• (i) Maximum torque: $\tau_{\text{max}} = m B = 8 \times 10^{-5} \times 0.5 = 4 \times 10^{-5}\text{ N\cdot m}$. • (ii) Net force on any closed loop in a uniform magnetic field is **zero** ($F_{\text{net}} = 0$).
Step 3: Torque at angle
Plane of coil makes $30^\circ$ with field $\implies$ angle $\theta$ between normal to plane (magnetic moment) and field is $\theta = 90^\circ - 30^\circ = 60^\circ$. $$\tau = m B \sin 60^\circ = 4 \times 10^{-5} \times \frac{\sqrt{3}}{2} \approx 3.46 \times 10^{-5}\text{ N\cdot m}$$
PYQ 52Case Study — 4M | CBSE 2023; 2024; 2025
Topic 6
52. Case Study — Force on a Current-Carrying Conductor: A straight wire carrying current I in a uniform magnetic field B experiences a force per unit length given by f = I(l⃗ × B⃗). The magnitude is F = BIL sin θ, where θ is the angle between current direction and field. The direction is given by Fleming's left-hand rule (FBI rule). This principle is used in electric motors. The force is maximum when the conductor is perpendicular to the field (θ = 90°) and zero when parallel (θ = 0°). (i) A wire of length 0.5 m carries a current of 3 A and is placed perpendicular to a field of 0.4 T. Find the force on the wire. (ii) Under what condition is the magnetic force on a current-carrying conductor zero? (iii) State Fleming's left-hand rule. (iv) A rectangular loop of dimensions 0.1 m × 0.05 m carries 2 A and is in a field of 0.3 T. Find the forces on each side when the loop lies in the plane of the field.
Final Answer: (i) 0.6 N, (ii) Parallel to field, (iii) FBI rule, (iv) Forces on opposite sides are equal and opposite.
Step 1: Part (i)
Given $L = 0.5\text{ m}$, $I = 3\text{ A}$, $B = 0.4\text{ T}$, $\theta = 90^\circ$: $$F = B I L \sin 90^\circ = 0.4 \times 3 \times 0.5 = 0.6\text{ N}$$
Step 2: Part (ii)
The force is zero when the conductor is aligned parallel or anti-parallel to the magnetic field vector ($\theta = 0^\circ$ or $180^\circ$).
Step 3: Part (iii)
Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. If the forefinger points in the direction of the **Magnetic Field** and the middle finger points along the **Current**, then the thumb points in the direction of the **Force (Motion)**.
Step 4: Part (iv)
The loop lies in the plane of the field: • Two sides of length $0.05\text{ m}$ are parallel to the field $\implies F = 0$. • Two sides of length $0.1\text{ m}$ are perpendicular to the field $\implies F = B I L = 0.3 \times 2 \times 0.1 = 0.06\text{ N}$ each, acting in opposite directions (out of paper and into paper).
TOPIC 7
Force Between Parallel Conductors & Ampere
8 Questions
PYQ 53MCQ — 1M | CBSE Recurring 2016–2026
Topic 7
53. Two long parallel wires separated by distance d carry currents I₁ and I₂ in the same direction. The force per unit length between them is:
(a) μ₀I₁I₂/(2πd), attractive
(b) μ₀I₁I₂/(2πd), repulsive
(c) μ₀I₁I₂/(4πd), attractive
(d) μ₀I₁I₂d/(2π), repulsive
Correct Answer: (a) μ₀I₁I₂/(2πd), attractive
Step 1: Force per unit length formula
The magnetic force per unit length between two long parallel wires is: $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$
Step 2: Force character
Parallel currents in the same direction **attract** each other. Correct option is (a).
PYQ 54MCQ — 1M | CBSE Recurring 2015–2025
Topic 7
54. Two long straight parallel conductors carry currents in opposite directions. The force between them is:
(a) Attractive
(b) Repulsive
(c) Zero
(d) Depends on magnitude of currents
Correct Answer: (b) Repulsive
Step 1: Opposite Currents behavior
Parallel conductors carrying currents in opposite directions (anti-parallel) **repel** each other. Correct option is (b).
PYQ 55MCQ — 1M | CBSE Recurring 2016–2025
Topic 7
55. One ampere is defined as that steady current which, when maintained in each of the two infinitely long parallel conductors of negligible cross-section separated by 1 m in vacuum, produces a force of:
(a) 4π × 10⁻⁷ N/m
(b) 2 × 10⁻⁷ N/m
(c) 1 N/m
(d) 10⁻⁷ N/m
Correct Answer: (b) 2 × 10⁻⁷ N/m
Step 1: Apply force formula
Using $I_1 = I_2 = 1\text{ A}$ and $d = 1\text{ m}$ in the force per unit length equation: $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2\pi \times 1} = 2 \times 10^{-7}\text{ N/m}$$ Correct option is (b).
56. Assertion (A): Two parallel current-carrying conductors with currents in the same direction attract each other. Reason (R): The magnetic field due to one conductor at the location of the other, combined with the force F = I(l × B) on the second conductor, gives a force directed towards the first conductor.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
Parallel currents in the same direction attract. Assertion (A) is true.
Step 2: Analyze Reason
If wire 1 carries current upward, it produces a magnetic field pointing into the page at wire 2. Using $F = I(\vec{L} \times \vec{B})$, wire 2 (current upward) experiences a force pointing left (towards wire 1). So Reason (R) is true.
Step 3: Check Explanation
Reason (R) describes the exact field-force interaction mechanism that results in attraction. Correct option is (a).
57. Two long straight parallel conductors carry steady currents in opposite directions. Using the concept of magnetic field, explain the nature of the force of interaction between them. Is it attractive or repulsive? Justify.
Final Answer: Repulsive force.
Step 1: Field lines configuration
• Wire 1 has current $I_1$ upward $\implies$ field at wire 2 points into page ($\otimes$). • Wire 2 has current $I_2$ downward.
Step 2: Apply Force direction rule
Apply Fleming's left-hand rule to wire 2: • Field: Into paper • Current: Downward • Force points **away from wire 1** (repulsion).
Step 3: Conclusion
Similarly, the force on wire 1 points away from wire 2. Hence, anti-parallel currents **repel** each other.
58. Derive an expression for the force per unit length between two long straight parallel current-carrying conductors separated by distance d. Hence define one ampere.
Final Answer: Derivation showing F/L = μ₀I₁I₂/(2πd).
Step 1: Compute Magnetic Field of Wire 1
Consider two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$. The magnetic field $B_1$ produced by wire 1 at the location of wire 2 is: $$B_1 = \frac{\mu_0 I_1}{2\pi d}$$
Step 2: Calculate force on Wire 2
The force experienced by length $L$ of wire 2 in this field $B_1$ is: $$F_2 = I_2 L B_1 \sin 90^\circ = I_2 L \left( \frac{\mu_0 I_1}{2\pi d} \right)$$
Step 3: Force per unit length
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$ By Newton's third law, an equal and opposite force acts on wire 1.
Step 4: Define Ampere
One ampere is that steady current which, when maintained in each of two infinitely long parallel conductors of negligible cross-section, separated by 1 meter in vacuum, produces a force of $2 \times 10^{-7}$ Newtons per meter of length between them.
59. (a) Two long straight parallel conductors carry currents I₁ = 3 A and I₂ = 5 A in the same direction and are separated by 20 cm. Find the force per unit length between them and state whether it is attractive or repulsive. (b) At what distance from the first wire is the resultant magnetic field zero?
Given $I_1 = 3\text{ A}$, $I_2 = 5\text{ A}$, $d = 20\text{ cm} = 0.2\text{ m}$: $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2 \times 10^{-7} \times 3 \times 5}{0.2} = 1.5 \times 10^{-5}\text{ N/m}$$ Since currents are parallel, the force is **attractive**.
Step 2: Balance of Magnetic Fields
Let the point of zero magnetic field be at distance $x$ from the first wire (between the wires since they carry parallel currents): $$B_1 = B_2 \implies \frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi(d - x)}$$
60. Case Study — Force Between Parallel Conductors and Definition of Ampere: Two long parallel current-carrying conductors exert forces on each other. The force per unit length is F/L = μ₀I₁I₂/(2πd). Parallel currents attract (force per unit length directed towards each other), while anti-parallel currents repel. This principle is used to define the SI unit of current, the Ampere: one ampere is that steady current which, when maintained in each of two infinitely long parallel conductors of negligible cross-section, separated by 1 m in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length between them. (i) Write the expression for force per unit length between two long parallel conductors. (ii) Two wires carry 5 A each in opposite directions and are 10 cm apart. Find the force per unit length between them. (iii) State whether the force is attractive or repulsive when currents are in the same direction. (iv) State the SI definition of one ampere.
Final Answer: (i) F/L = μ₀I₁I₂/(2πd), (ii) 5.0 × 10⁻⁵ N/m, (iii) Attractive, (iv) Standard definition.
When currents are in the same direction, the force is **attractive**.
Step 4: Part (iv)
One Ampere is defined as the current which, flowing through two infinitely long parallel wires separated by 1 m in vacuum, produces a force of $2 \times 10^{-7}\text{ N/m}$ between them.
PYQ 95LA — 5M | CBSE 2019; 2022; 2025; Recurring
Topic 7
95. (a) Derive an expression for the force per unit length between two infinitely long parallel conductors carrying currents I₁ and I₂, separated by distance d. State whether the force is attractive or repulsive in each case. (b) Hence define one ampere. (c) Two long parallel wires separated by 8 cm carry 5 A and 3 A in the same direction. Find the point between them where the resultant magnetic field is zero.
Final Answer: (c) $5\text{ cm}$ from the 5 A wire.
Step 1: Part (a) & (b) Explanations
Refer to Q58 for derivation and definition.
Step 2: Part (c) Field balance setup
Let point be at distance $x$ from the $5\text{ A}$ wire. Since currents are in the same direction, fields oppose between the wires: $$\frac{\mu_0(5)}{2\pi x} = \frac{\mu_0(3)}{2\pi(8 - x)} \implies \frac{5}{x} = \frac{3}{8 - x}$$
Step 3: Solve for x
$$5(8 - x) = 3x \implies 40 - 5x = 3x \implies 8x = 40 \implies x = 5\text{ cm}$$ The field is zero at a distance of **5 cm** from the 5 A wire (or 3 cm from the 3 A wire).
105. (a) Write the expression for force between two parallel current-carrying conductors. (b) Using this, define one ampere. (c) Show that parallel currents attract and anti-parallel currents repel, by considering the magnetic field of each wire and the force it exerts on the other. (d) Two long parallel wires are 4 cm apart. Wire A carries 3 A and wire B carries 5 A, both in the same direction. Find: (i) force per unit length, (ii) nature of force, (iii) the point between them where net B = 0.
Final Answer: (d) (i) $7.5 \times 10^{-5}\text{ N/m}$, (ii) Attractive, (iii) $1.5\text{ cm}$ from wire A.
Step 1: Part (a) & (b) & (c) Descriptions
• Expressions and definitions: See Q58. • Attraction/Repulsion justification: See Q56 and Q57.
Step 2: Part (d) (i) & (ii) Force calculations
Given $I_A = 3\text{ A}$, $I_B = 5\text{ A}$, $d = 0.04\text{ m}$: $$\frac{F}{L} = \frac{2 \times 10^{-7} \times 3 \times 5}{0.04} = 7.5 \times 10^{-5}\text{ N/m}$$ Nature of force: **Attractive** (since currents are in the same direction).
Step 3: Part (d) (iii) Find zero field point
Let the point be at distance $x$ from wire A: $$\frac{3}{x} = \frac{5}{4 - x} \implies 3(4 - x) = 5x \implies 12 - 3x = 5x \implies 8x = 12 \implies x = 1.5\text{ cm}$$
TOPIC 8
Torque on Current Loop
7 Questions
PYQ 61MCQ — 1M | CBSE Recurring 2016–2026
Topic 8
61. The torque experienced by a rectangular current loop (N turns, area A, current I) placed in a uniform magnetic field B at angle θ between the magnetic moment and B is:
(a) τ = NIAB cos θ
(b) τ = NIAB sin θ
(c) τ = NIB sin θ/A
(d) τ = NIA cos θ
Correct Answer: (b) τ = NIAB sin θ
Step 1: Torque formula
The torque on a magnetic dipole (current loop) in a magnetic field is: $$\vec{\tau} = \vec{m} \times \vec{B} \implies \tau = m B \sin\theta$$
Step 2: Substitute magnetic moment
Since magnetic moment is $m = N I A$: $$\tau = N I A B \sin\theta$$ Correct option is (b).
PYQ 62MCQ — 1M | CBSE Recurring 2017–2026
Topic 8
62. When is the torque on a current-carrying loop placed in a uniform magnetic field: (i) maximum, and (ii) zero?
(a) (i) Plane ∥ to field; (ii) Plane ⊥ to field
(b) (i) Plane ⊥ to field; (ii) Plane ∥ to field
(c) (i) Plane at 45° to field; (ii) Plane ∥ to field
(d) (i) Plane ⊥ to field; (ii) Plane at 45° to field
Correct Answer: (a) (i) Plane ∥ to field; (ii) Plane ⊥ to field
Step 1: Relate torque to plane angle
Let $\theta$ be the angle between field $\vec{B}$ and normal to loop plane. The torque is $\tau = NIAB\sin\theta$. • If loop plane is parallel to field, normal is perpendicular $\implies \theta = 90^\circ \implies \sin 90^\circ = 1 \implies$ maximum torque. • If loop plane is perpendicular to field, normal is parallel $\implies \theta = 0^\circ \implies \sin 0^\circ = 0 \implies$ zero torque.
Step 2: Conclusion
Option (a) is correct.
PYQ 63SA — 2M | CBSE Recurring 2015–2025
Topic 8
63. A rectangular coil ABCD of N turns, each of length l and breadth b, carries a current I in a uniform magnetic field B. Write the expression for: (i) force on each arm, and (ii) the torque acting on the coil when its plane makes angle α with B.
Final Answer: (ii) τ = NIAB cos α
Step 1: Part (i) Forces on arms
• The arms of length $l$ perpendicular to the field experience a force of $F = B I l$. These forces are equal and opposite, forming a couple. • The arms of length $b$ making angle with field experience forces of $F = B I b \cos\alpha$, which cancel each other along the axis.
Step 2: Part (ii) Torque expression
When the plane makes angle $\alpha$ with field, the normal makes angle $\theta = 90^\circ - \alpha$. $$\tau = N I A B \sin\theta = N I (lb) B \sin(90^\circ - \alpha) = N I A B \cos\alpha$$
64. Assertion (A): A current-carrying loop placed in a uniform magnetic field experiences zero net force but a non-zero torque. Reason (R): In a uniform field, the forces on opposite sides of the loop are equal and opposite (net force = 0), but they form a couple which produces a torque.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
For a closed loop, the vector sum of forces in a uniform field is $\oint I(d\vec{l} \times \vec{B}) = I(\oint d\vec{l}) \times \vec{B} = 0$. However, the forces act along different lines of action, causing rotation. Assertion (A) is true.
Step 2: Analyze Reason
The forces on opposite arms of a rectangular loop are equal in magnitude and opposite in direction. Since their lines of action do not coincide, they create a couple, producing torque. Reason (R) is true.
Step 3: Check Explanation
The couple formed by opposite equal-and-opposite forces explains the combination of zero net force and non-zero torque. Correct option is (a).
PYQ 65LA — 5M | CBSE Recurring 2015–2026
Topic 8
65. Derive an expression for the torque acting on a rectangular current-carrying loop of N turns, area A, carrying current I, placed in a uniform magnetic field B. Hence show that $\vec{\tau} = \vec{m} \times \vec{B}$, where $\vec{m}$ is the magnetic dipole moment.
Final Answer: Derivation of torque on a rectangular loop.
Step 1: Identify Forces on arms
Consider loop $ABCD$ with sides $AB = CD = a$ and $BC = DA = b$ carrying current $I$ in field $B$. Let the normal to plane make angle $\theta$ with $\vec{B}$. • Forces on arms $BC$ and $DA$ are equal, opposite, and act along the axis of rotation, so their net torque is zero. • Forces on arms $AB$ and $CD$ are $F_1 = F_2 = I a B$. They are opposite and act along parallel lines separated by perpendicular distance $d = b \sin\theta$.
Step 2: Calculate Torque of couple
Torque is force times perpendicular distance: $$\tau = F_1 (b \sin\theta) = (I a B) b \sin\theta = I (ab) B \sin\theta$$ Since area $A = a b$: $$\tau = I A B \sin\theta$$
Step 3: Vector Form
For $N$ turns, $\tau = N I A B \sin\theta$. Define magnetic moment vector $\vec{m} = N I A \hat{n}$: $$\vec{\tau} = \vec{m} \times \vec{B}$$
66. (a) Write the expression for the torque on a current-carrying loop in a uniform magnetic field. (b) A rectangular loop of length 20 cm and width 10 cm, carrying 5 A, is placed in a uniform field of 0.3 T with its plane parallel to the field. Find the torque acting on the loop. (c) What is the net mechanical force on the loop?
Final Answer: (b) $0.03\text{ N·m}$, (c) $0\text{ N}$
Step 1: Part (a) Expression
$$\vec{\tau} = \vec{m} \times \vec{B} \implies \tau = N I A B \sin\theta$$
Step 2: Part (b) Torque Calculation
Given $l = 0.2\text{ m}$, $w = 0.1\text{ m} \implies A = 0.2 \times 0.1 = 0.02\text{ m}^2$. Current $I = 5\text{ A}$, $B = 0.3\text{ T}$. Plane is parallel to field $\implies$ normal is perpendicular to field ($\theta = 90^\circ$): $$\tau = I A B \sin 90^\circ = 5 \times 0.02 \times 0.3 \times 1 = 0.03\text{ N\cdot m}$$
Step 3: Part (c) Net Force
Since the magnetic field is uniform, the forces on all segments of the closed loop cancel out vectorially. Thus, the net mechanical force on the loop is **zero**.
67. Two wires of equal lengths are bent into a square loop and a circular loop respectively. Both are placed in uniform magnetic field B and carry the same current I. Compare the torques on the two loops. Which is greater and why?
Final Answer: Circular loop torque is greater because of larger area.
Step 1: Setup area expressions
Let the length of both wires be $L$. • For square loop: Side $a = L/4 \implies A_{\text{sq}} = L^2/16$ • For circular loop: Radius $r = L/2\pi \implies A_{\text{cir}} = \pi r^2 = \frac{L^2}{4\pi}$
The circular loop experiences $\approx 1.27$ times the torque of the square loop because a circle has the maximum area for a given perimeter length.
TOPIC 9
Current Loop as Magnetic Dipole & Moment
9 Questions
PYQ 68MCQ — 1M | CBSE Recurring 2016–2026
Topic 9
68. The magnetic dipole moment of a current-carrying coil of N turns, each of area A, carrying current I, is:
(a) m = NIA
(b) m = NI/A
(c) m = NA/I
(d) m = I/NA
Correct Answer: (a) m = NIA
Step 1: Magnetic Dipole Moment Definition
The magnetic moment of a current loop is the product of current and the area enclosed: $$m = I A$$
Step 2: Extension to N turns
For a coil with $N$ turns, $m = N I A$. Correct option is (a).
PYQ 69MCQ — 1M | CBSE Recurring
Topic 9
69. The SI unit of magnetic dipole moment is:
(a) Tesla
(b) Joule/Tesla
(c) Ampere·m²
(d) Weber/m
Correct Answer: (c) Ampere·m²
Step 1: Dimension Check
From the magnetic moment formula $m = N I A$, the unit is Ampere (for current) times square meter (for area).
Step 2: Conclusion
Therefore, the SI unit is **$\text{A}\cdot\text{m}^2$** (Ampere-meter squared). Note: $\text{J/T}$ is also equivalent but $\text{A}\cdot\text{m}^2$ is the base SI unit. Correct option is (c).
PYQ 70MCQ — 1M | CBSE 2019; 2022; Recurring
Topic 9
70. Which of the following correctly gives the relation between magnetic moment $\vec{m}$ and angular momentum $\vec{L}$ of an electron revolving in an orbit of Bohr's atom?
(a) m⃗/L⃗ = −e/2mₑ
(b) m⃗/L⃗ = −e/mₑ
(c) m⃗/L⃗ = e/2mₑ
(d) m⃗/L⃗ = 2e/mₑ
Correct Answer: (a) m⃗/L⃗ = −e/2mₑ
Step 1: Revolving electron parameters
For a revolving electron of mass $m_e$ and charge $-e$: • Magnetic moment $m = I A = \frac{e}{T} (\pi r^2) = \frac{e v}{2\pi r} (\pi r^2) = \frac{evr}{2}$ • Angular momentum $L = m_e v r \implies vr = L/m_e$
Step 2: Relate variables vectorially
Substitute $vr$: $$m = \frac{e L}{2 m_e}$$ Since the electron is negatively charged, the magnetic moment vector is opposite to the orbital angular momentum vector (right-hand rules yield opposite directions): $$\frac{\vec{m}}{\vec{L}} = -\frac{e}{2 m_e}$$ Correct option is (a).
71. Assertion (A): A current-carrying loop behaves like a magnetic dipole. Reason (R): A current-carrying loop has two faces — one acts as north pole and the other as south pole, just like a bar magnet. Its magnetic dipole moment is m = NIA.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
A current loop produces a field identical to a tiny bar magnet at large distances, acting as a dipole. Assertion (A) is true.
Step 2: Analyze Reason
According to the direction of current, the face with counter-clockwise current is N pole and clockwise is S pole. The dipole moment is $NIA$. Reason (R) is true.
Step 3: Check Explanation
The existence of two polar faces and a defined magnetic moment is what makes the loop behave physically as a dipole. Correct option is (a).
PYQ 72SA — 2M | CBSE Recurring 2017–2025
Topic 9
72. Write an expression for the magnetic moment associated with a current I-carrying circular coil of radius r having N turns. Consider the above coil placed in YZ plane with its centre at origin. Which direction does its magnetic moment point?
Final Answer: m = N I (\pi r²). Direction is along +x or -x axis depending on current direction.
Step 1: Write Formula
The magnitude of the magnetic moment is: $$m = N I A = N I (\pi r^2)$$
Step 2: Find Direction in YZ plane
Since the coil lies in the YZ plane, the normal to the plane lies along the X-axis ($\hat{i}$ or $-\hat{i}$). • If current is counter-clockwise in YZ plane, $\vec{m}$ points along **$+x$-axis ($+\hat{i}$)**. • If clockwise, $\vec{m}$ points along **$-x$-axis ($-\hat{i}$)**.
PYQ 73SA — 3M | CBSE 2018; 2021; Recurring
Topic 9
73. An electron revolving around a nucleus of charge +Ze in an orbit of radius r with speed v. Show that it behaves like a tiny magnetic dipole. Hence derive an expression for the magnetic moment of the orbiting electron.
Final Answer: Derivation showing m = evr/2.
Step 1: Formulate equivalent current
An electron revolving in a circle of radius $r$ with speed $v$ completes one orbit in time $T = \frac{2\pi r}{v}$. The equivalent current is: $$I = \frac{e}{T} = \frac{e v}{2\pi r}$$
Step 2: Calculate Magnetic Moment
The magnetic dipole moment $m$ is current times area ($A = \pi r^2$): $$m = I A = \left( \frac{e v}{2\pi r} \right) \pi r^2 = \frac{e v r}{2}$$
Step 3: Conclusion
This shows that the orbiting charge creates a magnetic moment, behaving as a magnetic dipole.
PYQ 74Numerical — 3M | CBSE 2019; 2022; Recurring
Topic 9
74. A circular coil of 100 turns each of radius 8 cm carries a current of 0.4 A. (i) Calculate the magnetic dipole moment of the coil. (ii) Find the magnitude and direction of the torque when the coil is placed in a uniform magnetic field of 0.5 T with its plane parallel to the field.
Final Answer: (i) $0.804\text{ A·m}²$, (ii) $0.402\text{ N·m}$
Step 1: Calculate Area and Magnetic Moment
Given $N = 100$, $r = 0.08\text{ m}$, $I = 0.4\text{ A}$. Area $A = \pi r^2 = 3.1416 \times 0.08^2 \approx 0.0201\text{ m}^2$. $$m = N I A = 100 \times 0.4 \times 0.0201 = 0.804\text{ A\cdot m}^2$$
Step 2: Calculate Torque
Given $B = 0.5\text{ T}$, plane parallel to field $\implies \theta = 90^\circ$: $$\tau = m B \sin 90^\circ = 0.804 \times 0.5 \times 1 = 0.402\text{ N\cdot m}$$
Step 3: Direction
The torque acts in a direction perpendicular to both the magnetic moment vector and the magnetic field vector, tending to align the normal of the coil with the field.
PYQ 76Case Study — 4M | CBSE 2024; 2025; 2026
Topic 9
76. Case Study — Current Loop as Magnetic Dipole: A planar current loop (N turns, area A, current I) behaves exactly like a magnetic dipole with dipole moment m = NIA. The direction of m is given by the right-hand rule (curl fingers in direction of current; thumb points in direction of m). The torque on the dipole in a field B is τ = m × B = mB sin θ, which is identical to the torque on an electric dipole (p × E). The potential energy is U = −m⃗ · B⃗ = −mB cos θ. (i) Write the expression for magnetic moment of a coil. What is its SI unit? (ii) A coil of 50 turns, area 2 × 10⁻³ m², carries 2 A in B = 0.1 T. Find its magnetic moment and torque when plane is parallel to B. (iii) A current loop behaves like a bar magnet — which face acts as north pole and which as south? (iv) How is the magnetic moment of an orbiting electron related to its angular momentum?
Final Answer: (i) m = NIA, A·m², (ii) m = 0.2 A·m², τ = 0.02 N·m, (iii) CCW face is North, (iv) m/L = -e/2m.
Step 1: Part (i)
$$m = N I A \quad [\text{SI Unit: A}\cdot\text{m}^2]$$
Step 2: Part (ii)
• $m = 50 \times 2 \times 2 \times 10^{-3} = 0.2\text{ A\cdot m}^2$ • Plane parallel to field $\implies \theta = 90^\circ$: $\tau = m B \sin 90^\circ = 0.2 \times 0.1 = 0.02\text{ N\cdot m}$.
Step 3: Part (iii)
The face of the loop in which current flows **counter-clockwise** acts as the North pole, and the face with **clockwise** current acts as the South pole.
Step 4: Part (iv)
The ratio of magnetic moment to orbital angular momentum is the gyromagnetic ratio: $$\vec{m} = -\frac{e}{2 m_e} \vec{L}$$
PYQ 100LA — 5M | CBSE 2017; 2021; 2023; Recurring
Topic 9
100. (a) Define magnetic dipole moment of a current loop. (b) Derive the torque τ = m × B. Identify the positions of stable and unstable equilibrium. (c) An electron in Bohr's first orbit of hydrogen atom revolves with speed 2.18 × 10⁶ m/s. Calculate the magnetic moment associated with the orbital motion. [r₁ = 0.53 Å, e = 1.6 × 10⁻¹⁹ C]
Final Answer: (b) Stable at θ = 0°, Unstable at θ = 180°. (c) $9.24 \times 10^{-24}\text{ A·m}²$
Step 1: Part (a) & (b) Explanations
• Definition: See Q68. • Derivation: See Q65. • Potential Energy: $U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$. 1. **Stable equilibrium:** $\theta = 0^\circ$ ($U = -mB$ is minimum). 2. **Unstable equilibrium:** $\theta = 180^\circ$ ($U = +mB$ is maximum).
$$m = 0.8 \times 1.1554 \times 10^{-23} \approx 9.24 \times 10^{-24}\text{ A\cdot m}^2$$ Note: This value is very close to the Bohr Magneton ($\mu_B = 9.27 \times 10^{-24}\text{ A\cdot m}^2$).
75. (a) Derive the expression for the torque on a current loop. (b) Two moving coil meters M₁ and M₂ have: M₁: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T; M₂: R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T (spring constants identical). Find the ratio of (i) current sensitivity and (ii) voltage sensitivity of M₂ and M₁.
Final Answer: (b) (i) 1.4, (ii) 1.0
Step 1: Part (a) Derivation
Refer to Q65 for torque derivation.
Step 2: Part (b) Calculations
Refer to Q90 for step-by-step ratios calculations.
PYQ 77MCQ — 1M | CBSE Recurring 2017–2026
Topic 10
77. In a moving coil galvanometer, the deflection is proportional to:
(a) Voltage applied
(b) Square of current
(c) Current through the coil
(d) Resistance of the coil
Correct Answer: (c) Current through the coil
Step 1: Deflection formula
For a moving coil galvanometer, the restoring torque of the spring balances the magnetic torque at equilibrium: $$k \phi = N I A B \implies \phi = \left( \frac{N A B}{k} \right) I$$
Step 2: Conclusion
Since $N, A, B, k$ are constants, the angular deflection $\phi$ is **directly proportional to the current $I$** ($\phi \propto I$). Correct option is (c).
PYQ 78MCQ — 1M | CBSE Comptt. All India 2017; Recurring
Topic 10
78. The current sensitivity of a moving coil galvanometer is defined as:
(a) Deflection per unit voltage
(b) Deflection per unit current
(c) Current per unit deflection
(d) Charge per unit deflection
Correct Answer: (b) Deflection per unit current
Step 1: Define Current Sensitivity
Current sensitivity ($S_i$) is the deflection produced in the galvanometer per unit current passing through its coil: $$S_i = \frac{\phi}{I} = \frac{N A B}{k}$$ Correct option is (b).
PYQ 79MCQ — 1M | CBSE Recurring 2016–2026
Topic 10
79. A galvanometer of resistance G is converted into a voltmeter of range 0 to V volts by connecting a high resistance R in:
(a) Series with the galvanometer
(b) Parallel with the galvanometer
(c) Series with a battery
(d) Parallel with a battery
Correct Answer: (a) Series with the galvanometer
Step 1: Voltmeter Conversion principle
A voltmeter must have a very high resistance to ensure it draws negligible current when connected in parallel across a circuit component. This high resistance $R$ is achieved by connecting a large resistor in **series** with the galvanometer coil. Correct option is (a).
PYQ 80MCQ — 1M | CBSE Recurring 2015–2025
Topic 10
80. A galvanometer is converted into an ammeter of range 0 to I ampere by connecting a small shunt resistance S in:
(a) Series with the galvanometer
(b) Parallel with the galvanometer
(c) Between the galvanometer and battery
(d) In place of the galvanometer
Correct Answer: (b) Parallel with the galvanometer
Step 1: Ammeter Conversion principle
An ammeter must have a very low resistance so it does not alter the circuit current when inserted in series. This is done by bypassing most of the current through a small resistor (shunt $S$) connected in **parallel** across the galvanometer. Correct option is (b).
PYQ 81MCQ — 1M | CBSE Recurring 2019–2026
Topic 10
81. Which of the following correctly increases the current sensitivity of a moving coil galvanometer?
(a) Increasing the torsional constant of the spring
(b) Increasing the number of turns N
(c) Decreasing the area of the coil
(d) Increasing the resistance of the galvanometer
Correct Answer: (b) Increasing the number of turns N
Step 1: Check sensitivity formula
Current sensitivity is $S_i = \frac{N A B}{k}$.
Step 2: Compare options
• Increasing $k$ decreases $S_i$. • Increasing $N$ increases $S_i$. • Decreasing $A$ decreases $S_i$. • Resistance $R$ does not affect current sensitivity directly. Correct option is (b).
82. Assertion (A): Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Reason (R): Voltage sensitivity = Current sensitivity / Resistance. If current sensitivity is increased by increasing N (more turns), the resistance also increases, and voltage sensitivity may decrease or remain unchanged.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
If we double $N$, current sensitivity doubles, but since wire length doubles, the coil resistance $R$ also doubles. Thus, voltage sensitivity ($S_i/R$) remains unchanged. Assertion (A) is true.
Step 2: Analyze Reason
Voltage sensitivity is $S_v = \frac{NAB}{k R} = \frac{S_i}{R}$. Since $R$ grows with $N$, $S_v$ does not grow proportionally. Reason (R) is true.
Step 3: Check Explanation
The mathematical relationship $S_v = S_i / R$ explains why scaling $N$ increases $S_i$ but leaves $S_v$ unchanged. Correct option is (a).
83. Assertion (A): A voltmeter has a very high resistance while an ammeter has a very low resistance. Reason (R): A voltmeter must draw negligible current so it is connected in parallel and needs high resistance; an ammeter must not significantly alter the circuit current so it is connected in series and needs very low resistance.
(a) Both A and R true, R is correct explanation.
(b) Both A and R true, R is not correct explanation.
(c) A true, R false.
(d) Both false.
Correct Answer: (a) Both A and R true, R is correct explanation.
Step 1: Analyze Assertion
Ideal voltmeter resistance is $\infty$; ideal ammeter resistance is $0$. Assertion (A) is true.
Step 2: Analyze Reason
Voltmeter is placed in parallel to measure potential difference without bypassing current from the branch. Ammeter is in series to measure branch current directly. Reason (R) is true.
Step 3: Check Explanation
The measurement circuit configurations require these resistances to minimize load-effect measurement error. Correct option is (a).
PYQ 84SA — 2M | CBSE Comptt. All India 2017; Recurring 2016–2025
Topic 10
84. Define the term current sensitivity and voltage sensitivity of a moving coil galvanometer. Write the expression for each. Write their SI units.
Final Answer: Current sensitivity is deflection/current (rad/A). Voltage sensitivity is deflection/voltage (rad/V).
Step 1: Current Sensitivity Definition & SI Unit
Deflection produced per unit current: $S_i = \frac{\phi}{I} = \frac{N A B}{k}$. Unit: $\text{rad/A}$ (or div/A).
Step 2: Voltage Sensitivity Definition & SI Unit
Deflection produced per unit potential difference: $S_v = \frac{\phi}{V} = \frac{N A B}{k R}$ (where $R$ is coil resistance). Unit: $\text{rad/V}$ (or div/V).
PYQ 85SA — 2M | CBSE Recurring 2015–2024
Topic 10
85. State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer. Why should the spring/suspension wire have a low torsional constant?
Final Answer: Phosphor-bronze is used. Needs low torsional constant for high sensitivity.
Step 1: Suspension Wire Properties
Typically made of **Phosphor-Bronze** because: 1. **Low torsional constant ($k$):** Allows large deflection for small torque. 2. **High tensile strength:** Can support the weight of the coil without breaking. 3. **Non-magnetic nature:** Not affected by the magnet's field.
Step 2: Why Low Torsional Constant
Since current sensitivity $S_i = \frac{NAB}{k}$, reducing the torsional constant $k$ increases the sensitivity of the galvanometer, enabling detection of weaker currents.
PYQ 86SA — 3M | CBSE AI 2019; Recurring 2015–2025
Topic 10
86. Explain how you will convert a galvanometer into an ammeter to read a maximum current of I ampere. An ammeter is always connected in series with a circuit — why?
Final Answer: Connect a small shunt resistance S in parallel. Connected in series to measure full current.
Step 1: Shunt Connection
Connect a small resistance $S$ (called shunt) in parallel with the galvanometer of resistance $G$. Let $I_g$ be the current for full-scale deflection in the galvanometer. The remaining current $I - I_g$ flows through the shunt. Since they are in parallel: $$I_g G = (I - I_g) S \implies S = \frac{I_g G}{I - I_g}$$
Step 2: Why Series Connection
To measure the current flowing through a branch, the entire current must pass through the ammeter. Connecting it in series ensures this, and its extremely low resistance ($R_{\text{eq}} \approx S \to 0$) ensures it does not alter the original circuit current.
PYQ 87SA — 3M | CBSE Recurring 2015–2025
Topic 10
87. Explain with a circuit diagram how a galvanometer of resistance G (full scale deflection current $I_g$) can be converted into a voltmeter of range 0 to V volts. Derive the expression for the resistance to be connected. Why is a voltmeter connected in parallel?
Final Answer: Connect a high resistance R in series. R = V/Ig - G.
Step 1: Series Connection
Connect a large resistance $R$ in series with the galvanometer of resistance $G$. the total potential difference is: $$V = I_g (G + R)$$
Step 2: Derive R
$$\frac{V}{I_g} = G + R \implies R = \frac{V}{I_g} - G$$
Step 3: Why Parallel Connection
To measure the potential difference between two points, the voltmeter must be connected in parallel across those points so it experiences the same potential difference. Its high resistance prevents it from drawing significant current, leaving the circuit operation undisturbed.
PYQ 88Numerical — 3M | CBSE Recurring 2016–2025
Topic 10
88. A galvanometer has a resistance of 30 Ω and gives a full scale deflection for a current of 2 mA. (i) How will you convert it into an ammeter of range 0–0.3 A? (ii) How will you convert it into a voltmeter of range 0–3 V? Draw the circuit diagram for each and find the value of the resistance to be connected.
Final Answer: (i) Shunt S = 0.201 Ω in parallel, (ii) Series R = 1470 Ω in series.
Given $V = 3\text{ V}$: $$R = \frac{V}{I_g} - G = \frac{3}{0.002} - 30 = 1500 - 30 = 1470\ \Omega\text{ in series}$$
PYQ 89SA — 3M | CBSE Recurring 2017–2025
Topic 10
89. Define current sensitivity of a galvanometer. Increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. Justify this statement mathematically.
Final Answer: Doubling N doubles resistance, leaving voltage sensitivity unchanged.
Step 1: Write sensitivity formulas
$$S_i = \frac{N A B}{k}, \quad S_v = \frac{S_i}{R} = \frac{N A B}{k R}$$
Step 2: Math justification
Suppose we increase current sensitivity by doubling the number of turns ($N \to 2N$). This doubles $S_i$. However, doubling the turns requires doubling the length of the wire, which doubles the resistance of the coil ($R \to 2R$).
Step 3: Evaluate Sv
The new voltage sensitivity is: $$S_v' = \frac{2 S_i}{2 R} = S_v$$ Thus, the voltage sensitivity **remains unchanged**.
90. Two moving coil meters M₁ and M₂ have the following particulars: M₁: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T M₂: R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T (spring constants identical). Determine the ratio of (i) current sensitivity and (ii) voltage sensitivity of M₂ and M₁.
91. Case Study — Moving Coil Galvanometer: A moving coil galvanometer (MCG) works on the principle that a current-carrying coil in a uniform magnetic field experiences a torque. In equilibrium: τ_magnetic = τ_restoring, i.e., NIAB = kφ, where k is the torsional constant and φ is the deflection. So φ ∝ I. Current sensitivity Sᵢ = φ/I = NAB/k. Voltage sensitivity Sᵥ = φ/V = NAB/kR. The galvanometer can be converted: into an ammeter (shunt S in parallel), or into a voltmeter (high R in series). (i) Write the expression for current sensitivity and voltage sensitivity of an MCG. (ii) A galvanometer has R = 50 Ω, Iₘ = 1 mA. Find the shunt to convert it into ammeter of range 5 A. (iii) Find the series resistance to convert the above galvanometer into a voltmeter of range 10 V. (iv) Why should the sensitivity of an MCG be high? What factor limits maximum sensitivity?
Final Answer: (i) Si = NAB/k, Sv = NAB/kR, (ii) S = 0.01 Ω, (iii) R = 9950 Ω, (iv) Internal coil resistance increases with turns.
Step 1: Part (i)
$$S_i = \frac{N A B}{k}, \quad S_v = \frac{N A B}{k R}$$
Given $V = 10\text{ V}$: $$R = \frac{V}{I_g} - G = \frac{10}{0.001} - 50 = 10000 - 50 = 9950\ \Omega\text{ in series}$$
Step 4: Part (iv)
Sensitivity must be high to detect very weak signals. The factor limiting sensitivity is the coil's electrical resistance. Increasing $N$ to boost sensitivity adds wire length and mass, which increases resistance and moment of inertia, slowing response time and cancelling voltage sensitivity gains.
PYQ 94LA — 5M | CBSE 2018; 2020; 2023; Recurring
Topic 10
94. (a) Derive an expression for the torque on a rectangular N-turn coil carrying current I, placed in a uniform magnetic field B. Show τ = m × B. (b) Draw a labelled diagram of a moving coil galvanometer and explain its principle. (c) A galvanometer of G = 20 Ω, Iₘ = 5 mA is to be converted to (i) an ammeter of range 5 A, and (ii) a voltmeter of range 50 V. Find the values of resistance to be connected in each case.
Final Answer: (c) (i) Shunt S = 0.02 Ω in parallel, (ii) Series R = 9980 Ω in series.
Step 1: Part (a) & (b) Explanations
• Derivation: See Q65. • Principle: A current-carrying coil placed in a magnetic field experiences a torque, which is balanced by the restoring torque of a suspension spring.
Step 2: Part (c) (i) Ammeter conversion calculation
97. (a) Draw a labelled diagram of a moving coil galvanometer. Explain the principle, construction and working. (b) Write the expression for current sensitivity. Why is it important? (c) Define voltage sensitivity. Justify that increasing current sensitivity may not increase voltage sensitivity. (d) A galvanometer has R = 30 Ω, Iₘ = 3 mA. Find: (i) shunt for 0–3 A ammeter, (ii) series R for 0–30 V voltmeter.
Final Answer: (d) (i) S = 0.03 Ω, (ii) R = 9970 Ω.
Step 1: Part (a), (b), (c) Explanations
• Diagram & Working: Radial magnetic field ensures uniform torque $\tau = NIAB$ at all coil angles. Balanced by spring $k\phi$. • Current/Voltage sensitivity and justification: See Q84 and Q89.
104. (a) Explain the principle of a moving coil galvanometer with a neat labelled diagram. (b) Derive the equilibrium equation kφ = NIAB. (c) Define and derive the expressions for current sensitivity and voltage sensitivity. (d) A galvanometer has G = 25 Ω and Iₘ = 4 mA. It is to read up to 10 A as an ammeter and up to 25 V as a voltmeter. Find the required additional resistances and draw the circuit diagram in each case.
Final Answer: (d) S = 0.01 Ω in parallel, R = 6225 Ω in series.
Step 1: Part (a) & (b) & (c) Explanations
• Equilibrium: Deflection torque $\tau_d = N I A B \sin 90^\circ = N I A B$ (in radial field). Restoring torque $\tau_r = k\phi$. Equating them: $k\phi = N I A B$. • Sensitivity: See Q84.