Class 12 Physics · Original EZT practice
Moving Charges and Magnetism (Class 12 Chapter 4): summary and practice
A moving charge makes a magnetic field, and a magnetic field pushes back on a moving charge - this chapter turns that two-way link into formulas you can compute with, from the Lorentz force to the moving coil galvanometer. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Magnetic force and the Lorentz force
A charge q moving with velocity v in a magnetic field B feels F = q v B sin theta, with direction from the right-hand rule; the total Lorentz force adds the electric part qE. A current-carrying wire feels F = B I l sin theta.
2. Motion in a magnetic field
Perpendicular to B, a charged particle moves in a circle of radius r = mv/qB; with a parallel velocity component the path becomes a helix. The magnetic force does no work, so the speed stays constant.
3. Biot-Savart law
A small current element contributes dB proportional to I dl sin theta / r^2 at a field point, direction given by the right-hand rule. This builds the field of any circuit piece by piece.
4. Field of a circular current loop
On the axis of a loop of radius R carrying current I, the field at the centre is B = mu0 I / 2R; far away on the axis the loop behaves like a magnetic dipole.
5. Ampere's circuital law and the solenoid
Ampere's law says the line integral of B around a closed loop equals mu0 times the enclosed current. Applied to a long solenoid it gives B = mu0 n I inside, where n is turns per unit length.
6. Torque, dipole and the galvanometer
A current loop in a field feels torque = N I A B sin theta, which is why loops align with the field. The moving coil galvanometer uses this torque against a spring to measure current, with a radial field keeping the deflection proportional to current.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Compute the magnetic force on a charge
Choose the method: F = q v B sin theta.
EZT practice question
A charge of 1.6 x 10^-19 C moves at 2 x 10^6 m/s perpendicular to a field B = 0.5 T. Find the force on it.
Show the worked solution
- F = q v B sin theta, and sin 90 degrees = 1.
- = 1.6 x 10^-19 x 2 x 10^6 x 0.5.
Answer: F = 1.6 x 10^-13 N.
Check: Forces on single charges are tiny - the scientific notation answer is expected.
Watch for this: Do not drop sin theta: the force falls to zero when v is parallel to B.
Now try it yourself
Same charge and field, but velocity at 30 degrees to B - force?
Check your answer
8 x 10^-14 N.
sin 30 degrees = 0.5 halves the force.
Find the radius of circular motion
Choose the method: r = mv / qB.
EZT practice question
A proton (m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 C) moves at 10^6 m/s perpendicular to B = 0.1 T. Find the radius of its circular path.
Show the worked solution
- r = mv / qB.
- = (1.67 x 10^-27 x 10^6) / (1.6 x 10^-19 x 0.1) = 1.67 x 10^-21 / 1.6 x 10^-20.
Answer: r = 0.104 m.
Check: Faster or heavier particle means a bigger circle; stronger field means a tighter one.
Watch for this: Keep track of the powers of ten - they decide whether the answer is centimetres or metres.
Now try it yourself
Same proton at 2 x 10^6 m/s - new radius?
Check your answer
0.209 m.
r is proportional to v, so doubling v doubles r.
Compute the force on a current-carrying wire
Choose the method: F = B I l sin theta.
EZT practice question
A straight wire of length 0.4 m carries 5 A in a field B = 0.2 T. Find the force when the wire is at 90 degrees and at 30 degrees to the field.
Show the worked solution
- At 90 degrees: F = B I l = 0.2 x 5 x 0.4.
- At 30 degrees: multiply by sin 30 degrees = 0.5.
Answer: 0.4 N at 90 degrees and 0.2 N at 30 degrees.
Check: Maximum force comes at 90 degrees; the force falls as the wire aligns with the field.
Watch for this: Do not use 0.4 m in centimetres or forget the angle - both errors are common here.
Now try it yourself
Same wire at 90 degrees but carrying 10 A - force?
Check your answer
0.8 N.
F is proportional to I, so doubling the current doubles the force.
Find the field at the centre of a loop
Choose the method: B = mu0 I / 2R.
EZT practice question
A circular loop of radius 0.1 m carries 2 A. Find the magnetic field at its centre.
Show the worked solution
- B = mu0 I / 2R = (4 pi x 10^-7 x 2) / (2 x 0.1).
- = (8 pi x 10^-7) / 0.2.
Answer: B = 1.26 x 10^-5 T.
Check: Loop fields are of order 10^-5 T - much weaker than a bar magnet's surface field.
Watch for this: The centre formula has 2R in the denominator, not R squared - that form belongs to far-axis points.
Now try it yourself
Same loop with the current doubled - field at centre?
Check your answer
2.51 x 10^-5 T.
B is proportional to I.
Compute the field inside a solenoid
Choose the method: B = mu0 n I.
EZT practice question
A long solenoid has 1000 turns per metre and carries 0.5 A. Find the field inside it.
Show the worked solution
- B = mu0 n I.
- = 4 pi x 10^-7 x 1000 x 0.5.
Answer: B = 6.28 x 10^-4 T.
Check: Note the formula uses turns per metre, not the total number of turns.
Watch for this: Do not multiply by the solenoid's length - n already packs turns per unit length.
Now try it yourself
Same solenoid at 1 A - field inside?
Check your answer
1.26 x 10^-3 T.
B doubles with I.
Compute the torque on a current loop
Choose the method: torque = N I A B sin theta.
EZT practice question
A coil of 50 turns, area 0.01 m^2, carries 2 A in a field B = 0.2 T, with the plane of the coil parallel to the field. Find the torque.
Show the worked solution
- Plane parallel to B means theta = 90 degrees between the area vector and B, so sin theta = 1.
- torque = N I A B = 50 x 2 x 0.01 x 0.2.
Answer: 0.2 N m.
Check: Torque is maximum when the plane is parallel to the field and zero when the coil is aligned.
Watch for this: Theta is measured for the area vector (normal to the plane), not for the plane itself.
Now try it yourself
Same coil with the plane at 30 degrees to the field - torque?
Check your answer
0.173 N m.
The area vector is then at 60 degrees to B, and 0.2 x sin 60 degrees = 0.173 N m.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Force on a moving charge?
Show answer and reason
F = q v B sin theta.
Zero when v is parallel to B.
2. Radius of circular motion?
Show answer and reason
r = mv / qB.
The magnetic force supplies the centripetal force.
3. Field inside a solenoid?
Show answer and reason
B = mu0 n I.
From Ampere's circuital law.
4. Torque on a current loop?
Show answer and reason
N I A B sin theta.
Maximum when the plane is parallel to the field.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A proton at 2 x 10^6 m/s in B = 0.1 T: find the radius of its path.
- A 0.4 m wire carrying 10 A sits at 90 degrees to B = 0.2 T: find the force.
- A solenoid with 1000 turns per metre carries 1 A: find the field inside.
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
For cost context, see home tuition fees in Delhi NCR. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- Does a magnetic field change a charged particle's speed?
- No. The magnetic force is always perpendicular to the velocity, so it does no work and only changes the direction of motion, not the speed.
- What is the Lorentz force?
- The total force on a charge in electric and magnetic fields: F = qE + q(v x B). The magnetic part is q v B sin theta, directed by the right-hand rule.
- How does a moving coil galvanometer measure current?
- The current through its coil produces a torque N I A B in a radial magnetic field; a spring balances it, so the steady deflection is directly proportional to the current.
- Why is the field inside a long solenoid uniform?
- Ampere's circuital law applied to a rectangular loop through the solenoid shows the inside field is mu0 n I everywhere and the outside field is nearly zero.
- Are these the NCERT exercise questions for Moving Charges and Magnetism (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.