Class 12 Physics Notes
~9 min readMoving Charges and Magnetism Class 12 Physics Notes
Complete, exam-ready notes on moving charges and magnetism: Lorentz force, motion of a charge in a magnetic field, Biot-Savart law, Ampere's circuital law, solenoid and toroid, force between currents, circular loop, galvanometer and the cyclotron — written for CBSE boards, JEE and NEET revision.
What is the Lorentz force in one line?
The Lorentz force is the total force on a charged particle due to electric and magnetic fields: F = qE + q(v × B). The magnetic part acts perpendicular to both velocity and field, so it changes only the direction of motion, never the speed.
Lorentz Force and Motion in a Magnetic Field
Lorentz force
The magnetic force is always perpendicular to velocity, doing zero work. It causes a charged particle to move in a circular path of radius with angular frequency .
Helical path
Biot-Savart Law
Biot-Savart law
Gives the magnetic field produced by a small current element. For a long straight conductor, the field at distance a is .
- Field at the centre of a circular loop:
- .
- Field of a circular loop at a point on the axis:
- .
- Right-hand grip rule gives the direction (curl fingers along current).
Ampere's Circuital Law
Ampere's law
The line integral of the magnetic field around a closed loop equals μ₀ times the net current enclosed. It is used to find the field of symmetric current distributions.
- Solenoid field inside:
- (n = turns per unit length).
- Toroid field:
- .
- The field outside an ideal solenoid is zero.
Force Between Two Parallel Currents
Force between parallel wires
Parallel currents in the same direction attract; opposite directions repel. This defines the ampere: two parallel conductors 1 m apart carrying 1 A attract with a force of .
Moving Coil Galvanometer
Galvanometer deflection
Deflection is proportional to current, with N turns of area A in a field B and spring constant k. It detects small currents and is the basis of ammeters and voltmeters.
- To make an ammeter: connect a small shunt resistance in parallel.
- To make a voltmeter: connect a large series resistance.
- A radial magnetic field keeps the torque independent of the coil position, giving a linear scale.
Cyclotron
A cyclotron accelerates charged particles using a perpendicular magnetic field (which bends them in circles) and an alternating electric field across two dees. Since the angular frequency ω = qB/m is independent of speed, particles stay in resonance as they gain energy.
Relativistic limit
Solved Examples
Example: A proton moves with in a field perpendicular to its velocity. Find the radius ().
Solution: .
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q4.1–Q4.13)
- 4.1) A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
- 4.2) A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
- 4.3) A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
- 4.4) A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
- 4.5) What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
- 4.6) A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
- 4.7) Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
- 4.8) A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
- 4.9) A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
- 4.10) Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10⁻³ m², B1 = 0.25 T; R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10⁻³ m², B2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
- 4.11) In a chamber, a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s⁻¹ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10⁻¹⁹ C, me = 9.1 × 10⁻³¹ kg)
- 4.12) In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
- 4.13) (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Answers at a glance
- 1Q4.1 → B = 3.14 × 10⁻⁴ T (direction perpendicular to the coil's plane).
- 2Q4.2 → B = 3.5 × 10⁻⁵ T.
- 3Q4.3 → B = 4.0 × 10⁻⁶ T, directed vertically upward.
- 4Q4.4 → B = 1.2 × 10⁻⁵ T, directed south (horizontally toward the south).
- 5Q4.5 → Force per unit length = 0.6 N m⁻¹.
- 6Q4.6 → F = 8.1 × 10⁻² N.
- 7Q4.7 → F = 2 × 10⁻⁶ N, attractive.
- 8Q4.8 → B ≈ 2.5 × 10⁻² T.
- 9Q4.9 → τ = 0.96 N m.
- 10Q4.10 → (a) Current sensitivity ratio M2:M1 = 1.4; (b) Voltage sensitivity ratio M2:M1 = 1.
- 11Q4.11 → The magnetic force is always ⊥ v and acts as centripetal force, giving a circle; r = 4.2 × 10⁻² m = 4.2 cm.
- 12Q4.12 → ν ≈ 18 MHz; independent of speed because T = 2πm/eB contains no v.
- 13Q4.13 → (a) Counter-torque = 3.1 N m. (b) No — torque depends only on the enclosed area, not the coil's shape.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1State Biot–Savart law in vector form.
- 2What is the force on a stationary charge in a uniform magnetic field?
- 3State the Lorentz force law.
- 4On what principle does the moving coil galvanometer operate?
2 Mark Questions
- 1A long straight wire carries a current of 5 A. Find the magnetic field at a point 10 cm from the wire. (Ans: B = μ₀I/2πd = 10⁻⁵ T)
- 2A proton of charge 1.6 × 10⁻¹⁹ C moves at 2 × 10⁶ m/s perpendicular to a field of 0.5 T. Find the radius of its path given mass 1.67 × 10⁻²⁷ kg. (Ans: r = mv/qB ≈ 4.2 cm)
- 3A coil of 100 turns and radius 5 cm carries 2 A. Find the field at its centre. (Ans: B = μ₀NI/2R ≈ 2.5 × 10⁻³ T)
3 Mark Questions
- 1Using Biot–Savart law, derive the magnetic field at a point on the axis of a circular current loop. (Ans: B = μ₀IR²/[2(R² + x²)^(3/2)])
- 2Derive the magnetic field inside a long solenoid assuming it to be closely wound and tightly packed. (Ans: B = μ₀nI)
- 3Derive the expression for the force on a current-carrying conductor placed at an angle θ to a uniform magnetic field. (Ans: F = IBl sin θ)
5 Mark Questions
- 1Describe the principle, construction and working of a cyclotron, derive the expression for the cyclotron frequency and show that ions gain energy across the dees. (Ans: ω = qB/m)
- 2Derive the magnetic field at the centre of a current-carrying circular loop using Biot–Savart law and obtain the magnetic moment of a current loop as m = NIA.
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all six topics: Lorentz force and motion in a magnetic field; Biot-Savart law; Ampere's circuital law; force between two parallel currents; moving coil galvanometer; and cyclotron. Arranged topic by topic.
Higher-difficulty set with multi-topic integration problems (charged-particle motion, field calculations, force between currents, galvanometer conversion, cyclotron design) and full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Lorentz force
Circular radius
Angular frequency
Straight wire field
Loop centre field
Solenoid field
Force between wires
Galvanometer
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- Magnetic force does no work — it changes direction, not speed.
- Cyclotron frequency ω = qB/m is independent of velocity.
- Parallel currents attract; anti-parallel repel.
- Ampere's law: ∮B·dl = μ₀I_enc.
- Biot-Savart gives dB; integrate for extended wires and loops.
FAQ
Frequently asked questions
Why does a magnetic field do no work on a moving charge?
The magnetic force F = q(v × B) is always perpendicular to the velocity, so its dot product with displacement is zero — it changes only direction, never speed or energy.
What is the difference between Biot-Savart law and Ampere's law?
Biot-Savart law computes the field from a current element by integration; Ampere's law uses symmetry (∮B·dl = μ₀I_enc) to find fields of highly symmetric distributions faster.
How is a galvanometer converted into an ammeter?
By connecting a small shunt resistance in parallel, which diverts most of the current, letting the galvanometer measure a small calibrated fraction of the total current.
Why are parallel currents attracting?
Each current produces a magnetic field that exerts a Lorentz force on the other. By the right-hand rule, same-direction currents produce an attractive force between them.
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