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Moving Charges and Magnetism Class 12 Case Based Questions 2026-27
Moving Charges and Magnetism Class 12 Case Based Questions
8 original case-study sets with answers, calculations and concept-based reasoning.
Designed for current CBSE 2026–27 preparation. These are original practice questions, not official CBSE questions or predictions.
Current-syllabus focus: magnetic field and Oersted's experiment, Biot–Savart law and circular loop, Ampere's law for an infinitely long straight wire, Lorentz force, force on current-carrying conductors, force between parallel currents, torque and magnetic dipole, and moving-coil galvanometer. Straight-solenoid treatment is qualitative in the current syllabus.
How to Use These Case Based Questions
Read the passage first without looking at the answers. For numerical parts, write the formula, substitute values in SI units, and state the final answer with its unit. For direction questions, use the appropriate right-hand rule or the vector direction of v × B.
Case Study 1 — Charged Particle in a Uniform Magnetic Field
A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. The magnetic force is always perpendicular to the instantaneous velocity. The particle therefore follows a circular path with constant speed.
(a) Why is the path circular?
Answer: The magnetic force is perpendicular to the velocity and acts as the centripetal force required for circular motion.
(b) Write the radius of the circular path.
Answer:
r = mv / (|q|B)
(c) If the speed is doubled while q, m and B remain unchanged, what happens to r?
Answer: The radius doubles because r is directly proportional to speed.
(d) Does the magnetic field change the kinetic energy of the particle?
Answer: No. Magnetic force is perpendicular to velocity, so it does no work. Hence the speed and kinetic energy remain unchanged.
Case Study 2 — Crossed Electric and Magnetic Fields
A charged beam travels through mutually perpendicular electric and magnetic fields. At one particular speed, the electric force and magnetic force are equal in magnitude and opposite in direction, so the beam passes through undeflected.
(a) Write the magnitudes of the electric and magnetic forces when the velocity is perpendicular to B.
Answer:
FE = |q|E FB = |q|vB
(b) Obtain the condition for zero deflection.
Answer: For zero net force, |q|E = |q|vB, therefore v = E/B.
(c) If E = 3 × 104 N/C and B = 0.20 T, find the undeflected speed.
Answer:
v = E/B = (3 × 104)/(0.20) = 1.5 × 105 m/s
(d) What happens if the particle travels faster than the selected speed, with E and B unchanged?
Answer: The magnetic force becomes greater than the electric force, so the particle is deflected in the direction of the resultant force.
Case Study 3 — Oersted Experiment and a Long Straight Wire
A straight conductor carries a steady current. A compass placed near the conductor shows a deflection, demonstrating that electric current produces a magnetic field. Around a long straight conductor, the field lines are concentric circles.
(a) What does the compass deflection demonstrate?
Answer: It demonstrates that a current-carrying conductor produces a magnetic field around it.
(b) Which rule gives the direction of the magnetic field around the conductor?
Answer: The right-hand thumb rule.
(c) If the distance from the long wire is doubled, what happens to B?
Answer: It becomes half because B ∝ 1/r.
(d) A wire carries 6 A. Find B at 5 cm from it in vacuum.
Answer:
B = μ0I/(2πr) = (4π × 10−7 × 6)/(2π × 0.05) = 2.4 × 10−5 T
Case Study 4 — Biot–Savart Law and a Circular Coil
A circular coil of N turns, radius R and current I produces a magnetic field at its centre. For a circular coil in vacuum, the magnitude of the field at the centre is proportional to NI and inversely proportional to R.
(a) Write the magnetic field at the centre of an N-turn circular coil.
Answer:
B = μ0NI/(2R)
(b) What happens to B if the current is doubled?
Answer: B doubles, because B is directly proportional to I.
(c) What happens to B if the radius is doubled while N and I remain unchanged?
Answer: B becomes half, because B is inversely proportional to R.
(d) For N = 20, R = 0.10 m and I = 2 A, calculate B at the centre.
Answer:
B = (4π × 10−7 × 20 × 2)/(2 × 0.10) = 2.51 × 10−4 T (approximately)
Case Study 5 — Force on a Current-Carrying Conductor
A straight conductor of length L carrying current I is placed in a uniform magnetic field B. The magnetic force depends on the angle between the conductor/current direction and the magnetic field.
(a) Write the magnitude of the magnetic force on the conductor.
Answer:
F = BIL sin θ
(b) At what angle is the force maximum?
Answer: At θ = 90°, because sin 90° = 1.
(c) At what angle is the force zero, apart from the trivial case I = 0 or B = 0?
Answer: At θ = 0° or 180°, when the conductor is parallel or antiparallel to the magnetic field.
(d) A 0.40 T field acts on a 3 A wire of length 0.50 m at 90°. Find the force.
Answer:
F = BIL = 0.40 × 3 × 0.50 = 0.60 N
Case Study 6 — Force Between Parallel Current-Carrying Conductors
Two long parallel conductors carry currents I1 and I2 and are separated by distance d. The magnetic interaction between them depends on the product of the currents and varies inversely with their separation.
(a) What happens when the two currents flow in the same direction?
Answer: The conductors attract each other.
(b) What happens when the currents flow in opposite directions?
Answer: The conductors repel each other.
(c) Write the force per unit length between the conductors in vacuum.
Answer:
F/L = μ0I1I2/(2πd)
(d) For I1 = 4 A, I2 = 6 A and d = 0.30 m, find F/L.
Answer:
F/L = (4π × 10−7 × 4 × 6)/(2π × 0.30) = 1.6 × 10−5 N/m
Case Study 7 — Torque and Magnetic Dipole Moment
A current-carrying coil placed in a uniform magnetic field behaves like a magnetic dipole. For an N-turn coil of area A carrying current I, its magnetic dipole moment has magnitude NIA and the torque depends on the angle between the magnetic moment and the field.
(a) Write the magnitude of the magnetic dipole moment of the coil.
Answer:
m = NIA
(b) Write the torque on the coil in a uniform magnetic field.
Answer:
τ = mB sin θ = NIAB sin θ
(c) At what angle is the torque maximum?
Answer: θ = 90°.
(d) A 100-turn coil has area 0.02 m², carries 0.50 A and is in a 0.30 T field. Find its maximum torque.
Answer:
τmax = NIAB = 100 × 0.50 × 0.02 × 0.30 = 0.30 N m
Case Study 8 — Moving-Coil Galvanometer and Its Conversion
A moving-coil galvanometer works on the torque experienced by a current-carrying coil in a magnetic field. At equilibrium, magnetic torque is balanced by restoring torque. Its current sensitivity depends on the number of turns, coil area, magnetic field and torsional constant.
(a) Write the expression for current sensitivity.
Answer:
SI = θ/I = NAB/k
(b) How can current sensitivity be increased if the other quantities are unchanged?
Answer: Increase N, A or B, or decrease the torsional constant k.
(c) A galvanometer has resistance G = 100 Ω and full-scale current Ig = 1 mA. Find the shunt needed to convert it into a 1 A ammeter.
Answer:
S = IgG/(I − Ig) = (0.001 × 100)/(1 − 0.001) ≈ 0.100 Ω
(d) Using the same galvanometer, what series resistance is required to make a 10 V voltmeter?
Answer:
R = V/Ig − G = 10/0.001 − 100 = 9900 Ω
Quick Revision Map
| Case | Main concept tested | Skill |
|---|---|---|
| 1 | Lorentz force and charged-particle motion | Concept + formula |
| 2 | Crossed electric and magnetic fields | Derivation + numerical |
| 3 | Oersted experiment and long straight wire | Concept + field calculation |
| 4 | Biot–Savart law and circular coil | Proportional reasoning + numerical |
| 5 | Force on current-carrying conductor | Formula + application |
| 6 | Parallel current-carrying conductors | Direction + numerical |
| 7 | Magnetic dipole and torque | Formula + numerical |
| 8 | Moving-coil galvanometer | Sensitivity + instrument conversion |
Exam trap: Do not confuse the angle in F = BIL sin θ (between current direction and B) with the angle in τ = mB sin θ (between magnetic moment and B). For a charged particle, always use the angle between v and B in F = |q|vB sin θ.
Continue Your Chapter 4 Preparation
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Moving Charges and Magnetism Numericals
Official reference: For the latest syllabus and assessment documents, use the official CBSE Academic website. This practice page is independently prepared for revision and should not be treated as an official CBSE publication.
Important: These case-based questions are original practice material. They are designed to reflect the concepts and skills of the current syllabus, but they are not leaked, copied, official, or guaranteed board-exam questions.
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