Moving Charges and Magnetism Class 12 Notes: Chapter Overview

In these Moving Charges and Magnetism Class 12 notes, the chapter is Chapter 4 of NCERT Class 12 Physics (Part 1). It shows that a moving charge produces a magnetic field and that a magnetic field exerts a force on a moving charge. This link between electricity and magnetism explains devices such as electric motors and the galvanometer.

The chapter covers:

• Magnetic force on a moving charge, and the Lorentz force in electric and magnetic fields
• Motion of a charged particle in a magnetic field
• Biot–Savart law and the magnetic field of a circular current loop
• Ampere’s circuital law, with the straight wire, solenoid and toroid
• Force on a current-carrying conductor, and the force between two parallel currents
• Torque on a current loop, magnetic dipole moment and the moving coil galvanometer

Syllabus note: the cyclotron was removed from the CBSE board syllabus in the earlier rationalisation, and the treatment of the solenoid and toroid is qualitative. Check the latest CBSE syllabus for boards, and the NTA and NMC syllabi for JEE and NEET.

Key Formulas in Moving Charges and Magnetism Class 12 Notes

Motion of a Charged Particle in a Magnetic Field: Moving Charges and Magnetism Class 12 Notes

The magnetic force on a charge is always perpendicular to its velocity, so it does no work and does not change the speed. It only changes the direction.

• v perpendicular to B: the charge moves in a circle of radius r = mv/(qB).
• v parallel to B: the force is zero, and the charge moves in a straight line.
• v at an angle to B: the charge moves in a helix.
• Velocity selector: in crossed E and B fields, charges with v = E/B pass undeflected.

Cyclotron (JEE): the frequency f = qB/(2πm) does not depend on the speed of the particle, so a fixed-frequency alternating voltage can accelerate it each half cycle. The maximum kinetic energy is q²B²R²/(2m), where R is the radius of the dees.

Magnetic Field Due to Current in Moving Charges and Magnetism Class 12 Notes

Use the right-hand thumb rule: if the thumb points along the current, the curled fingers give the direction of the magnetic field around a straight wire.

Force on Current-Carrying Conductors and Parallel Wires in Moving Charges and Magnetism Class 12 Notes

A conductor of length L carrying current I in a field B experiences a force F = BIL sin θ. Its direction comes from the Fleming’s left-hand rule.

Two long parallel wires with currents in the same direction attract each other, and with currents in opposite directions they repel each other. The force per unit length is μ₀I₁I₂/(2πd).

Note on the ampere: NCERT explains the ampere through the force between two parallel wires. Since the 2019 revision of the SI, the ampere is defined by fixing the value of the elementary charge, but the parallel-wire relation is still used in exam problems.

Torque on a Current Loop and the Galvanometer in Moving Charges and Magnetism Class 12 Notes

A loop with N turns, area A and current I in a field B experiences a torque τ = NIAB sin θ. Its magnetic dipole moment is m = NIA.

In a moving coil galvanometer, the magnetic torque on the coil is balanced by a spring torque, so the deflection is φ = (NAB/k) I. The current sensitivity is φ/I = NAB/k, and the voltage sensitivity is φ/V = NAB/(kR). A radial magnetic field and a soft iron core keep the deflection proportional to the current.

• Ammeter: a small shunt resistance S = I_g G/(I − I_g) is connected in parallel with the galvanometer.
• Voltmeter: a large resistance R = V/I_g − G is connected in series with the galvanometer.

Solved Examples from Moving Charges and Magnetism Class 12 Notes

Example 1 (circular path): An electron (m = 9.1 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C) moves at 2 × 10⁶ m/s perpendicular to a magnetic field of 0.01 T. Find the radius of its path.

r = mv/(qB) = (9.1 × 10⁻³¹ × 2 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.01) ≈ 1.1 × 10⁻³ m (about 1.1 mm).

Example 2 (circular loop): Find the magnetic field at the centre of a circular loop of radius 10 cm that carries a current of 5 A.

B = μ₀I/(2R) = (4π × 10⁻⁷ × 5)/(2 × 0.1) ≈ 3.14 × 10⁻⁵ T.

Example 3 (parallel wires): Two long parallel wires carry currents of 10 A and 5 A in the same direction, 10 cm apart. Find the force per unit length.

F/L = μ₀I₁I₂/(2πd) = (2 × 10⁻⁷ × 10 × 5)/0.1 = 1 × 10⁻⁴ N/m, and the wires attract each other.

Example 4 (solenoid): A long solenoid has 500 turns per metre and carries 2 A. Find the field inside.

B = μ₀nI = 4π × 10⁻⁷ × 500 × 2 ≈ 1.26 × 10⁻³ T.

Example 5 (galvanometer): A galvanometer has a resistance of 50 Ω and gives full-scale deflection for 5 mA. Find the shunt needed for a 0–1 A ammeter, and the series resistance for a 0–10 V voltmeter.

Shunt S = I_gG/(I − I_g) = (0.005 × 50)/(1 − 0.005) ≈ 0.25 Ω. Series resistance R = V/I_g − G = 10/0.005 − 50 = 1950 Ω.

Moving Charges and Magnetism Class 12 Notes

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

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Common Mistakes in Moving Charges and Magnetism Class 12 Notes Problems

What Do CBSE, JEE and NEET Ask from Moving Charges and Magnetism Class 12 Notes?

The number of questions varies every year, so use previous year questions to see the trend.

How to Study Moving Charges and Magnetism Class 12 Notes

1. Read the NCERT chapter first, and solve the examples and exercises. eSaral’s NCERT Solutions for Class 12 Physics can help you check your answers.
2. Revise from the formula table and the handwritten notes above.
3. Practise the right-hand and left-hand rules from the Moving Charges and Magnetism Class 12 notes, because many mistakes come from directions.
4. Revise the chapter with eSaral’s Class 12 Physics revision notes before the exam. You can also download the textbook from the NCERT Class 12 books page.