Class 12 Physics Chapter 4 Important Questions (2026-27)
Important Questions for Class 12 Physics Chapter 4 – Moving Charges and Magnetism
Looking for Class 12 Physics Chapter 4 important questions? This page covers Moving Charges and Magnetism with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on the Biot–Savart law, Ampere's law, the Lorentz force and the galvanometer. Attempt them after finishing the NCERT chapter.
Very Short Answer Questions (1 Mark)
Magnetic Force and Direction
Question 1
What is the direction of the force acting on a charged particle \(q\), moving with a velocity \(\vec{v}\) in a uniform magnetic field \(\vec{B}\)?
Question 2
An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field?
Question 3
A beam of \(\alpha\)-particles projected along the +x-axis experiences a force due to a magnetic field along the +y-axis. What is the direction of the magnetic field?
Question 4
A beam of electrons projected along the +x-axis experiences a force due to a magnetic field along the +y-axis. What is the direction of the magnetic field?
Question 5
A beam of protons projected along the +x-axis experiences a force due to a magnetic field along the –y-axis. What is the direction of the magnetic field?
Question 6
Write the expression in vector form for the magnetic force \(\vec{F}\) acting on a charged particle moving with velocity \(\vec{v}\) in the presence of a magnetic field \(\vec{B}\).
Question 7
Write the expression in vector form for the Lorentz magnetic force \(\vec{F}\) due to a charge moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\). What is the direction of the magnetic force?
The force is normal to both the directions of velocity \(\vec{v}\) and magnetic field \(\vec{B}\).
Question 8
Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields.
Question 9
A particle of mass \(m\) and charge \(q\) moving with velocity \(v\) enters a region of uniform magnetic field at right angles to the direction of its motion. How does its kinetic energy get affected?
Question 10
Depict the trajectory of a charged particle moving with velocity \(\vec{v}\) as it enters a uniform magnetic field perpendicular to the direction of its motion.
Parallel Currents and Ampere
Question 11
Using the concept of force between two infinitely long parallel current-carrying conductors, define one ampere of current.
Galvanometer and Magnetic Moment
Question 12
Why should the spring/suspension wire in a moving coil galvanometer have a low torsional constant?
Question 13
Write the underlying principle of a moving coil galvanometer.
Question 14
A coil of area \(A\) carrying a steady current \(I\) has a magnetic moment \(\vec{m}\) associated with it. Write the relation between \(\vec{m}\), \(I\) and \(A\) in vector form.
Solenoid and Toroid
Question 15
Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?
Short Answer Questions (2–3 Marks)
Biot–Savart Law
Question 16
State Biot–Savart law. 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 element \(d\vec{l}\) at point P situated at a distance \(\vec{r}\) from the element.
\(dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}\)
Vector form: \(d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{l} \times \vec{r})}{r^3}\)
The direction is given by the right-hand screw rule, perpendicular to the plane of \(d\vec{l}\) and \(\vec{r}\).
Question 17
(i) State Biot–Savart law in vector form expressing the magnetic field \(d\vec{B}\) due to an element \(d\vec{l}\) carrying current \(I\) at a distance \(\vec{r}\) from the element. (ii) Write the expression for the magnitude of the magnetic field at the centre of a circular loop of radius \(r\) carrying a steady current \(I\). Draw the field lines due to the current loop.
(ii) At the centre of a circular loop: \(B = \dfrac{\mu_0 NI}{2r}\). For a single turn \(N = 1\): \(B = \dfrac{\mu_0 I}{2r}\).
Field lines: circular loops concentric with the coil; inside the loop, the lines are almost parallel to the axis.
Question 18
(a) State Biot–Savart law and express this law in vector form. (b) Two identical circular coils, P and Q each of radius R, carrying currents 1 A and \(\sqrt{3}\) A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils.
(b) \(B_P = \dfrac{\mu_0 I_P}{2R} = \dfrac{\mu_0 (1)}{2R}\), \(B_Q = \dfrac{\mu_0 I_Q}{2R} = \dfrac{\mu_0 (\sqrt{3})}{2R}\)
Since the two coils are perpendicular, \(B = \sqrt{B_P^2 + B_Q^2} = \dfrac{\mu_0}{2R}\sqrt{1 + 3} = \dfrac{\mu_0}{2R} \times 2 =\) \(\dfrac{\mu_0}{R}\)
Direction: \(\tan\theta = \dfrac{B_Q}{B_P} = \sqrt{3} \Rightarrow \theta = 60^\circ\) from \(B_P\) towards \(B_Q\).
Question 19
Use Biot–Savart law to derive the expression for the magnetic field on the axis of a current-carrying circular loop of radius R. Draw the magnetic field lines due to a circular wire carrying current I.
For an element \(dl\) at \(C\), the distance \(PC = r = \sqrt{a^2 + x^2}\). By Biot–Savart law:
\(dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{(a^2 + x^2)}\)
The components perpendicular to the axis cancel; parallel components add up. Integrating over the loop:
\(B = \dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}\)
At the centre, \(x = 0\): \(B = \dfrac{\mu_0 I}{2a}\).
Field lines: circular loops about the wire; inside the loop, the lines are parallel to the axis.
Question 20
(a) Draw a schematic sketch of a moving coil galvanometer and describe briefly its working. (b) "Increasing the current sensitivity of a galvanometer does not necessarily increase the voltage sensitivity." Justify this statement.
(b) Current sensitivity \(I_s = \dfrac{\phi}{I} = \dfrac{NAB}{k}\). Voltage sensitivity \(V_s = \dfrac{\phi}{V} = \dfrac{\phi}{IR} = \dfrac{I_s}{R}\).
Increasing \(I_s\) (by increasing N or A) also tends to increase \(R\) (since more turns means more resistance), so \(V_s\) may stay the same or even decrease. Hence increasing current sensitivity does not necessarily increase voltage sensitivity.
Question 21
Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer. Can a galvanometer as such be used for measuring the current? Explain.
Working: Deflecting torque \(= NAIB\), restoring torque \(= k\phi\). In equilibrium, \(NABI = k\phi\), so \(\phi \propto I\).
(i) Radial field: The magnetic field is always in the plane of the coil (normal to the area vector), so the torque is maximum (\(\sin\theta = 1\)) in every position.
(ii) Soft iron core: Makes the field radial and stronger, increasing torque and sensitivity.
Current sensitivity: \(I_s = \dfrac{\phi}{I} = \dfrac{NAB}{k}\).
No, a galvanometer cannot be used directly to measure current. It has a high resistance, which would change the circuit current, and it gives full-scale deflection for very small currents, so it could be damaged by large currents.
Question 22
Define the current sensitivity of a galvanometer. Write its SI unit.
Question 23
State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer cannot be used as such to measure current in a given circuit. Name any two factors on which the current sensitivity of a galvanometer depends.
Two reasons for not using as ammeter: (i) It is very sensitive (small full-scale deflection) — large currents can damage it. (ii) Its large resistance would change the circuit current if connected in series.
Two factors for current sensitivity: (a) Number of turns N, (b) Area A (and magnetic field B, torsional constant k).
Question 24
Explain the principle and working of a cyclotron with the help of a schematic diagram. Write the expression for cyclotron frequency.
Working: A positive ion enters the gap between the two dees and is accelerated toward the negative dee. Inside the dee, it experiences no electric field (shielded), but the perpendicular magnetic field makes it move in a circular path. When it exits, the polarity of the dees has reversed and it is again accelerated. Repeating this, the particle follows a spiral path and gains high energy.
Cyclotron frequency: \(f = \dfrac{qB}{2\pi m}\), independent of speed and energy.
Question 25
State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.
Question 26
(i) Obtain the expression for the cyclotron frequency. (ii) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer.
Time period: \(T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}\). Frequency: \(f = \dfrac{1}{T} = \dfrac{qB}{2\pi m}\).
(ii) The mass of a deuteron and a proton is different. Since \(f \propto \dfrac{1}{m}\), the same oscillator frequency cannot accelerate both particles. They cannot be accelerated with the same oscillator frequency.
Question 27
Write the expression for the magnetic force \(\vec{F}\) acting on a charged particle \(q\) moving with velocity \(\vec{v}\) in the presence of a magnetic field \(\vec{B}\) in vector form. Show that no work is done and no change in the magnitude of the velocity of the particle is produced by this force. Hence define the unit of magnetic field.
The force is always perpendicular to the instantaneous velocity \(\vec{v}\), hence perpendicular to the displacement \(d\vec{s}\). Work done \(W = \vec{F} \cdot d\vec{s} = 0\). So kinetic energy and speed remain constant.
Unit of B: One tesla is the magnetic field at a point where a charge of one coulomb moving with velocity \(1\ \text{m/s}\) perpendicular to \(\vec{B}\) experiences a force of one newton.
Question 28
Write the expression for Lorentz magnetic force on a particle of charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\). Show that no work is done by this force on the charged particle.
The force is always perpendicular to the velocity, so the work done \(W = \int \vec{F} \cdot d\vec{s} = 0\). Hence no work is done, and the kinetic energy of the particle remains unchanged.
Parallel Currents and Force Between Wires
Question 29
A steady current \(I_1\) flows through a long straight wire. Another wire carrying steady current \(I_2\) in the same direction is kept close and parallel to the first wire. Show with a diagram how the magnetic field due to the current \(I_1\) exerts a magnetic force on the second wire. Write the expression for this force.
Force on length \(L\) of wire 2: \(F = I_2 L B_1 = \dfrac{\mu_0 I_1 I_2 L}{2\pi d}\).
Since the currents are in the same direction, the force is attractive. The force per unit length: \(\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}\).
Question 30
Derive the expression for force per unit length between two long straight parallel current-carrying conductors. Hence define one ampere.
Field due to \(I_1\) at wire 2: \(B_1 = \dfrac{\mu_0 I_1}{2\pi d}\).
Force on length \(L\) of wire 2: \(F = I_2 L B_1 = \dfrac{\mu_0 I_1 I_2 L}{2\pi d}\). So \(\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}\). Force is attractive.
Ampere: One ampere is that current which, if maintained in two infinitely long parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of \(2 \times 10^{-7}\ \text{N}\) per metre on each conductor.
Question 31
Two long straight parallel conductors carry steady currents \(I_1\) and \(I_2\) separated by a distance \(d\). If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.
Ampere: As in Question 30.
Solenoid and Toroid
Question 32
Using Ampere's circuital law, obtain an expression for the magnetic field along the axis of a current-carrying solenoid of length \(l\) and having \(N\) turns.
\(\oint \vec{B} \cdot d\vec{l} = Bx\) (only the side along the axis contributes).
Current enclosed by the loop: number of turns in length \(x\) is \(nx\), so \(I_{enc} = nx I_0\).
By Ampere's law: \(Bx = \mu_0 (nx I_0) \Rightarrow\) \(B = \mu_0 n I_0\)
Question 33
(a) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases. (b) How is the magnetic field inside a given solenoid made strong?
Field lines: In a solenoid, the field is along the axis (uniform inside, diverge at the ends). In a toroid, the field lines are circular closed loops confined within the core.
(b) The magnetic field inside a solenoid can be made stronger by (i) inserting a soft iron core, and (ii) increasing the current.
Question 34
(a) How is a toroid different from a solenoid? (b) Use Ampere's circuital law to obtain the magnetic field inside a toroid. (c) Show that in an ideal toroid, the magnetic field (i) inside the toroid and (ii) outside the toroid in the open space is zero.
(b) Consider an Amperian loop (region II) of radius r inside the toroid. By symmetry, B is along the loop. \(B \cdot 2\pi r = \mu_0 (n \cdot 2\pi r) I\) where \(n\) is turns per unit length.
\(B = \mu_0 n I\) (inside the toroid)
(c) (i) Inside an ideal toroid (in the open space inside the ring), current enclosed by a small circular loop is zero, so \(B = 0\).
(ii) Outside the toroid (in the open space around the ring), current going out equals current coming in — net current enclosed is zero — so \(B = 0\).
Question 35
Draw the magnetic field lines due to a current passing through a long solenoid. Use Ampere's circuital law to obtain the expression for the magnetic field due to the current \(I\) in a long solenoid having \(n\) turns per unit length.
\(\oint \vec{B} \cdot d\vec{l} = Bx\). Current enclosed = \(nx I\).
\(Bx = \mu_0 (nxI) \Rightarrow\) \(B = \mu_0 nI\)
Question 36
A long straight wire of circular cross-section of radius \(a\) carries a steady current \(I\). The current is uniformly distributed across the cross-section. Apply Ampere's circuital law to calculate the magnetic field at a point distance \(r\) from the axis for (i) \(r < a\) and (ii) \(r > a\).
By Ampere's law: \(B \cdot 2\pi r = \mu_0 I \dfrac{r^2}{a^2} \Rightarrow\) \(B = \dfrac{\mu_0 I r}{2\pi a^2}\)
(ii) For \(r > a\): Enclosed current = \(I\).
\(B \cdot 2\pi r = \mu_0 I \Rightarrow\) \(B = \dfrac{\mu_0 I}{2\pi r}\)
Question 37
A long straight wire of circular cross-section (radius = \(a\)) carries a current \(I\) which is uniformly distributed across the cross-section. Use Ampere's circuital law to calculate the magnetic field \(B(r)\) at a point at distance \(r < a\) and \(r > a\) from its axis. Draw a graph showing the dependence of \(B(r)\) on \(r\).
For \(r < a\): \(B = \dfrac{\mu_0 I r}{2\pi a^2}\) (linear in r).
For \(r > a\): \(B = \dfrac{\mu_0 I}{2\pi r}\) (inverse in r).
Graph: B increases linearly from 0 at r=0 to max \(\dfrac{\mu_0 I}{2\pi a}\) at r = a, then decreases as 1/r for r > a.
Question 38
A toroidal solenoid of mean radius 20 cm has 4000 turns of wire wound on a ferromagnetic core of relative permeability 800. Calculate the magnetic field in the core for a current of 3 A passing through the coil. How does the field change when this core is replaced by a core of Bismuth?
\(B = \mu_r \mu_0 n I = 800 \times 4\pi \times 10^{-7} \times 3183 \times 3\)
\(B \approx 800 \times (4\pi \times 10^{-7}) \times 9549 \approx 800 \times 1.2 \times 10^{-2} \approx\) 9.6 T
Bismuth is diamagnetic (\(\mu_r < 1\)), so the field would be very much reduced (slightly less than the field in vacuum).
Magnetic Force and Torque on Current Loops
Question 39
A rectangular coil of sides \(a\) and \(b\) carrying a current \(I\) is subjected to a uniform magnetic field \(\vec{B}\) acting perpendicular to its plane. Obtain the expression for the torque acting on it.
On DA (length b, current in): \(F_1 = IbB\) inward.
On BC: \(F_2 = IbB\) outward. \(F_1 = F_2\) — net force is zero, but they form a couple.
Perpendicular distance between them = a. Torque = \(F_1 \times a = IbB \cdot a = IAB\) where A = ab.
If coil has n turns and angle between \(\vec{A}\) and \(\vec{B}\) is θ: \(\vec{\tau} = nI\vec{A} \times \vec{B}\), magnitude \(\tau = nIAB\sin\theta\).
When θ = 90°, \(\tau_{max} = nIAB\). When θ = 0° or 180°, \(\tau = 0\).
Question 40
Derive the expression for the torque \(\tau\) acting on a rectangular current loop of area A placed in a uniform magnetic field B. Show that \(\vec{\tau} = \vec{m} \times \vec{B}\) where \(\vec{m}\) is the magnetic moment of the current loop given by \(\vec{m} = I\vec{A}\).
Question 41
A square-shaped plane coil of area 100 cm² of 200 turns carries a steady current of 5 A. It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the direction of the field. In which orientation will the coil be in stable equilibrium?
If the plane makes 60° with B, then the normal to the plane makes θ = (90° – 60°) = 30° with B.
\(\tau = NIAB\sin\theta = 200 \times 5 \times 10^{-2} \times 0.2 \times \sin 30^\circ = 200 \times 0.05 \times 0.2 \times 0.5 =\) 1 N·m
Stable equilibrium: when the plane of the coil is parallel to \(\vec{B}\) (i.e., \(\vec{A}\) is parallel to \(\vec{B}\)).
Question 42
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.
\(qE = qvB \Rightarrow v = \dfrac{E}{B}\)
Only particles with this speed pass through undeflected — a velocity selector.
Question 43
A charge q of mass m is moving with a velocity v at right angles to a uniform magnetic field B. Deduce the expression for the radius of the circular path it describes.
So \(r = \dfrac{mv}{qB}\)
Question 44
A proton and an alpha particle having the same kinetic energy are passed through a region of uniform magnetic field, acting normal to the plane of the paper, and travel in circular paths. Deduce the ratio of the radii of the circular paths described by them.
\(r = \dfrac{mv}{qB} = \dfrac{m}{qB}\sqrt{\dfrac{2K}{m}} = \dfrac{\sqrt{2mK}}{qB}\)
So \(r \propto \dfrac{\sqrt{m}}{q}\). For proton (\(m, e\)) and alpha particle (\(4m, 2e\)):
\(\dfrac{r_p}{r_\alpha} = \dfrac{\sqrt{m}/e}{\sqrt{4m}/(2e)} = \dfrac{\sqrt{m}/e}{2\sqrt{m}/(2e)} = 1\). Ratio = 1 : 1.
Question 45
Derive the expression for the magnetic moment \((\vec{\mu})\) of an electron revolving around the nucleus in terms of its angular momentum. What is the direction of the magnetic moment of the electron with respect to its angular momentum?
Magnetic moment: \(\mu = IA = \dfrac{ev}{2\pi r} \cdot \pi r^2 = \dfrac{evr}{2}\).
Orbital angular momentum: \(L = mvr \Rightarrow vr = \dfrac{L}{m}\).
So \(\mu = \dfrac{e L}{2m_e}\), vector form \(\vec{\mu} = -\dfrac{e}{2m_e}\vec{L}\)
The negative sign indicates that the magnetic moment is opposite in direction to the angular momentum (since electron has negative charge).
Question 46
An electron moves around the nucleus in a hydrogen atom of radius 0.51 Å, with a velocity of \(2 \times 10^5\ \text{m/s}\). Calculate (i) the equivalent current due to orbital motion of electron, (ii) the magnetic field produced at the centre of the nucleus, (iii) the magnetic moment associated with the electron.
(i) \(T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 5.1 \times 10^{-11}}{2 \times 10^5} \approx 1.6 \times 10^{-15}\ \text{s}\).
\(I = \dfrac{e}{T} = \dfrac{1.6 \times 10^{-19}}{1.6 \times 10^{-15}} \approx\) \(1.0 \times 10^{-4}\ \text{A}\)
(ii) \(B = \dfrac{\mu_0 I}{2r} = \dfrac{4\pi \times 10^{-7} \times 10^{-4}}{2 \times 5.1 \times 10^{-11}} \approx\) 12.3 T
(iii) \(\mu = IA = 10^{-4} \times \pi (5.1 \times 10^{-11})^2 \approx\) \(8.2 \times 10^{-25}\ \text{A}\cdot\text{m}^2\)
Question 47
Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus.
Question 48
An electron of mass \(m_e\) revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated with it is expressed as \(\vec{\mu} = -\dfrac{e}{2m_e}\vec{L}\) where L is the orbital angular momentum of the electron. Give the significance of the negative sign.
\(I = \dfrac{e}{T} = \dfrac{ev}{2\pi r}\), \(\mu = IA = \dfrac{evr}{2}\), \(L = m_evr\).
\(\vec{\mu} = -\dfrac{e}{2m_e}\vec{L}\). The negative sign shows that \(\vec{\mu}\) and \(\vec{L}\) are in opposite directions (due to the electron's negative charge).
Motion in Magnetic Field
Question 49
(a) Write the expression for the force \(\vec{F}\) acting on a particle of mass m and charge q moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\). Under what conditions will it move in (i) a circular path and (ii) a helical path? (b) Show that the kinetic energy of a particle moving in a magnetic field remains constant.
(i) Circular path: when \(\vec{v}\) is perpendicular to \(\vec{B}\).
(ii) Helical path: when \(\vec{v}\) has a component parallel to \(\vec{B}\) and a component perpendicular to it.
(b) The magnetic force is perpendicular to velocity at every instant, so the work done by it is zero. Hence speed and hence kinetic energy remain constant.
Question 50
(a) If \(\vec{v}\) is perpendicular to \(\vec{B}\), show that a charged particle describes a circular path with angular frequency \(\omega = \dfrac{qB}{m}\). (b) If the velocity \(\vec{v}\) has a component parallel to the magnetic field \(\vec{B}\), trace the path described by the particle. Justify.
Angular velocity: \(\omega = \dfrac{v}{r} = \dfrac{qB}{m}\).
(b) If \(\vec{v}\) has a parallel component \(v_{||}\), it stays constant (no force along B), and the perpendicular component causes circular motion. The particle traces a helical path. Pitch: \(p = v_{||}T = \dfrac{2\pi m v_{||}}{qB}\).
Question 51
A proton and a deuteron, each moving with velocity \(\vec{v}\), enter simultaneously into a region of magnetic field \(\vec{B}\) acting normal to the direction of velocity. Trace their trajectories establishing the relationship between the two.
\(r_p : r_d = \dfrac{m}{e} : \dfrac{2m}{e} = 1 : 2\)
Both follow circular paths; the deuteron path has twice the radius. They both complete one revolution in the same time period \(T = \dfrac{2\pi m}{qB} \propto \dfrac{m}{q}\); but here \(T_p : T_d = \dfrac{m}{e} : \dfrac{2m}{e} = 1 : 2\).
Question 52
A uniform magnetic field \(\vec{B}\) is set up along the positive x-axis. A particle of charge q and mass m moving with velocity v enters the field at the origin in the XY plane such that it has velocity components both along and perpendicular to the magnetic field. Trace, giving reason, the trajectory followed by the particle. Find the expression for the distance moved by the particle along the magnetic field in one rotation.
Pitch \(p = v_\parallel T = v_\parallel \cdot \dfrac{2\pi m}{qB}\) = \(\dfrac{2\pi m v_\parallel}{qB}\)
Question 53
A neutron, an electron and an alpha particle, moving with equal velocities, enter a uniform magnetic field going into the plane of the paper. Trace their paths in the field and justify your answer.
- Neutron: charge zero → no force → moves in a straight line undeviated.
- Electron: negative charge → deflects in a circle, in the direction determined by the left-hand rule for negative charge.
- Alpha particle: positive charge → deflects in the opposite direction (opposite curvature to electron).
Radius \(r = \dfrac{mv}{qB}\) — electron has smaller radius due to its smaller mass; alpha particle has larger radius due to its larger mass-to-charge ratio.
Long Answer Questions (5 Marks)
Biot–Savart Law and Circular Coils
Question 54
(a) Using Biot–Savart's 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. (b) Two small identical circular coils marked 1, 2 carry equal currents and are placed with their geometric axes perpendicular to each other. Derive an expression for the resultant magnetic field at O.
\(dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin 90^\circ}{R^2} = \dfrac{\mu_0 I\,dl}{4\pi R^2}\)
Integrating over the whole loop (length \(2\pi R\)): \(B = \dfrac{\mu_0 I}{4\pi R^2}(2\pi R) = \dfrac{\mu_0 I}{2R}\). For N turns: \(B = \dfrac{\mu_0 NI}{2R}\)
(b) Both coils produce magnetic fields \(B_1 = B_2 = \dfrac{\mu_0 NI}{2R}\) at O, but along perpendicular axes.
Resultant: \(B = \sqrt{B_1^2 + B_2^2} = \sqrt{2} \cdot \dfrac{\mu_0 NI}{2R} =\) \(\dfrac{\mu_0 NI}{2R}\sqrt{2}\), directed at 45° to each field.
Question 55
Use Biot–Savart law to derive the expression for the magnetic field at a point on the axis of a current-carrying circular loop of radius R. Hence find the magnetic field at the centre of the loop.
\(B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\)
At the centre (\(x = 0\)): \(B = \dfrac{\mu_0 I}{2R}\)
Question 56
Find the magnetic field at a point on the axis of a circular coil carrying current and hence find the magnetic field at the centre of the circular coil.
Galvanometer
Question 57
(a) Draw a schematic sketch of a moving coil galvanometer and describe briefly its working. (b) "Increasing the current sensitivity of a galvanometer does not necessarily increase the voltage sensitivity." Justify this statement.
Cyclotron
Question 58
Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, stating clearly the function of the electric and magnetic field applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle.
Working: The particle is accelerated by the electric field across the gap between dees. Inside a dee, it moves in a circular path (shielded from the electric field by the metallic dee). Each time it crosses the gap, the field reverses, giving it another push. The particle spirals outward, gaining energy.
Function of fields: The electric field provides acceleration; the magnetic field makes the particle move in circles (keeps it confined in the dees).
Period: Centripetal force \(=\dfrac{mv^2}{r} = qvB \Rightarrow r = \dfrac{mv}{qB}\). Period \(T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}\) — independent of velocity/energy.
Parallel Currents and Torque
Question 59
Three long straight parallel wires are kept as shown. The wire (3) carries a current I. (i) The direction of flow of current I in wire (3) is such that the net force on wire (1), due to the other two wires, is zero. (ii) By reversing the direction of I, the net force on wire (2), due to the other two wires, becomes zero. What will be the directions of current I in the two cases? Also obtain the relation between the magnitudes of currents \(I_1\), \(I_2\) and I.
(ii) When the direction of I is reversed (flowing upwards), the net force on wire (2) becomes zero, meaning forces due to wires 1 and 3 on wire 2 are equal and opposite.
Relations (assuming wires 1 and 2 are separated by distance d, and wires 2 and 3 also by d): for case (ii): force from 1 on 2 \(= \dfrac{\mu_0 I_1 I_2}{2\pi d}\) and force from 3 on 2 \(= \dfrac{\mu_0 I I_2}{2\pi d}\). For zero net force: \(I = I_1\).
Key Formulas to Remember
| Concept | Formula |
|---|---|
| Lorentz force | \(\vec{F} = q(\vec{v} \times \vec{B})\) |
| Biot–Savart law | \(d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{l} \times \vec{r})}{r^3}\) |
| B at centre of circular coil | \(B = \dfrac{\mu_0 NI}{2R}\) |
| B on the axis of circular coil | \(B = \dfrac{\mu_0 IR^2}{2(R^2 + x^2)^{3/2}}\) |
| B inside a long solenoid | \(B = \mu_0 nI\) |
| B inside a toroid | \(B = \mu_0 nI\) |
| B due to long straight wire | \(B = \dfrac{\mu_0 I}{2\pi r}\) |
| B inside a thick wire (\(r < a\)) | \(B = \dfrac{\mu_0 I r}{2\pi a^2}\) |
| Force per unit length (parallel wires) | \(\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}\) |
| Torque on a current loop | \(\vec{\tau} = \vec{m} \times \vec{B}\), \(\tau = NIAB\sin\theta\) |
| Magnetic moment | \(m = NIA\) |
| Radius of circular path | \(r = \dfrac{mv}{qB}\) |
| Cyclotron frequency | \(f = \dfrac{qB}{2\pi m}\), \(T = \dfrac{2\pi m}{qB}\) |
| Velocity selector | \(v = \dfrac{E}{B}\) (when \(\vec{E} \perp \vec{B} \perp \vec{v}\)) |
| Magnetic moment of electron | \(\vec{\mu} = -\dfrac{e}{2m_e}\vec{L}\) |
| Galvanometer sensitivity | \(I_s = \dfrac{NAB}{k}\), \(V_s = \dfrac{I_s}{R}\) |
FAQs
Are these questions enough for CBSE Class 12 Physics Chapter 4?
Yes. These questions cover every major topic — Lorentz force, Biot–Savart law, magnetic field of a solenoid and toroid, force between parallel currents, torque on a current loop, moving coil galvanometer and cyclotron — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.
Which topics are most important for board exams?
Biot–Savart law and its applications, magnetic field of a circular coil and solenoid, force between parallel current-carrying conductors, torque on a current loop, moving coil galvanometer (including sensitivity), and cyclotron (principle, working and frequency) are consistently high-weightage topics.
How should I prepare for numericals in this chapter?
Practise circular-path radius and period problems (r = mv/qB), magnetic field calculations using Biot–Savart and Ampere's laws, torque on current loops, force per unit length between parallel wires, and galvanometer sensitivity problems. Focus on understanding the direction of the force using the right-hand rule or Fleming's left-hand rule.