Moving charges and magnetism for NEET

The NEET unit on magnetic effects of current and magnetism covers two NCERT chapters: moving charges and magnetism, and magnetism and matter. The first carries most of the numericals. A charged particle circling in a field, the field at the centre of a loop, a galvanometer turned into an ammeter: these are standard, and each has a formula you can apply in under a minute.

The second chapter is lighter and more conceptual. It is mainly about bar magnets as dipoles and the three kinds of magnetic material, which makes it a good place to pick up quick marks.

Force on a moving charge

The Lorentz force on a charge q moving with velocity v through fields E and B is

F = q(E + v × B)

The magnetic part has magnitude qvB sin θ and is perpendicular to both v and B. Because it is always perpendicular to the velocity, a magnetic force does no work: it changes the direction of motion but never the speed or kinetic energy.

Motion in a uniform magnetic field:

• If v is perpendicular to B, the charge moves in a circle with radius r = mv/(qB) = p/(qB) = √(2mK)/(qB).
• The period is T = 2πm/(qB), which does not depend on the speed or the radius. A faster particle moves in a bigger circle and takes the same time.
• If v makes an angle θ with B, the path is a helix. The pitch (distance moved along B per turn) is v cos θ × T.

Crossed fields (velocity selector): if E and B are perpendicular to each other and to v, the electric and magnetic forces cancel when v = E/B. Only particles with that speed pass through undeflected.

Force on a current and between currents

A straight conductor of length l carrying current I in a field B feels F = IlB sin θ.

Two long parallel wires a distance d apart exert a force per unit length

F/l = μ₀I₁I₂/(2πd), with μ₀ = 4π × 10⁻⁷ T m A⁻¹.

Currents in the same direction attract; opposite currents repel. This force was used to define the ampere before the 2019 SI revision.

Magnetic fields of currents

The Biot–Savart law gives the field of a small current element: dB = (μ₀/4π) I dl sin θ / r². Ampère's circuital law gives the same results more quickly when the geometry is symmetric: the line integral of B around a closed loop equals μ₀ times the current enclosed.

The arc formula gives μ₀I/(2R) for a full circle (θ = 2π) and μ₀I/(4R) for a semicircle, which is a useful check.

Torque on a current loop and the galvanometer

A current loop behaves as a magnetic dipole with moment m = NIA, directed by the right-hand rule. In a uniform field it feels a torque

τ = m × B, of magnitude NIAB sin θ

where θ is the angle between the normal to the loop and B. The torque is largest when the plane of the loop is parallel to B.

In a moving coil galvanometer, a radial field keeps the plane of the coil parallel to B, so the torque is NIAB at every position. It is balanced by a spring torque kφ, giving φ = (NAB/k)I. The deflection is proportional to the current.

• Current sensitivity = φ/I = NAB/k.
• Voltage sensitivity = φ/V = NAB/(kG), where G is the coil resistance.

Doubling the number of turns doubles the current sensitivity but also roughly doubles G, so the voltage sensitivity may not improve.

Converting a galvanometer (resistance G, full-scale current I_g):

An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.

Magnetism and matter

A bar magnet behaves like a solenoid: its field lines leave the north pole, enter the south pole, and form closed loops through the magnet. As a dipole of moment m:

• Field on the axis at distance r: (μ₀/4π)(2m/r³).
• Field on the equatorial line: (μ₀/4π)(m/r³).
• Torque in a uniform field: τ = mB sin θ; potential energy U = −mB cos θ.

These mirror the electric dipole results in electrostatics for NEET.

A superconductor expels the magnetic field completely (the Meissner effect). It is a perfect diamagnet.

Worked numericals

Example 1: radius of a proton's path

A proton (m = 1.67 × 10⁻²⁷ kg) moves at 2 × 10⁶ m s⁻¹ perpendicular to a 0.1 T field.

• r = mv/(qB) = (1.67 × 10⁻²⁷ × 2 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.1) = (3.34 × 10⁻²¹)/(1.6 × 10⁻²⁰).
• r ≈ 0.21 m, about 21 cm.

Example 2: proton and alpha particle

A proton and an alpha particle (mass 4m, charge 2e) enter the same field perpendicularly. Compare their radii when (a) they have the same kinetic energy, and (b) they are accelerated through the same voltage.

• (a) r = √(2mK)/(qB) ∝ √m/q. For the alpha particle, √4/2 = 1. Radii are equal.
• (b) K = qV, so r = √(2mV/q)/B ∝ √(m/q). For the alpha particle, √(4/2) = √2. Its radius is √2 times the proton's.

Example 3: field at the centre of a coil

A coil of 100 turns and radius 0.1 m carries 5 A.

• B = μ₀NI/(2R) = (4π × 10⁻⁷ × 100 × 5)/0.2 = 4π × 10⁻⁷ × 2,500 = π × 10⁻³ T.
• B ≈ 3.14 × 10⁻³ T.

Example 4: converting a galvanometer both ways

A galvanometer has G = 99 Ω and full-scale current 1 mA.

• Ammeter of range 100 mA: S = (1 × 99)/(100 − 1) = 1 Ω in parallel.
• Voltmeter of range 10 V: R = 10/0.001 − 99 = 10,000 − 99 = 9,901 Ω in series.

Example 5: force between parallel wires

Two long parallel wires 10 cm apart each carry 10 A in the same direction.

• F/l = (4π × 10⁻⁷ × 10 × 10)/(2π × 0.1) = (2 × 10⁻⁷ × 100)/0.1 = 2 × 10⁻⁴ N m⁻¹, attractive.

Practice MCQs

1. The current in a circular loop is doubled and its radius is doubled. The field at the centre: (a) is unchanged (b) doubles (c) halves (d) becomes 4 times
2. A charged particle's speed is doubled in the same uniform field. The radius of its circular path: (a) halves (b) is unchanged (c) doubles (d) becomes 4 times
3. In the same situation, the period of revolution: (a) halves (b) is unchanged (c) doubles (d) becomes 4 times
4. The work done by a magnetic field on a moving charge is: (a) qvB (b) qvB sin θ (c) depends on the path (d) zero
5. The turns per metre of a solenoid are doubled and the current is halved. The field inside: (a) doubles (b) is unchanged (c) halves (d) becomes 4 times
6. A coil of 50 turns and area 0.01 m² carries 2 A in a 0.2 T field, with its plane parallel to the field. The torque is: (a) zero (b) 0.02 N m (c) 0.2 N m (d) 2 N m
7. A substance with a small negative susceptibility is: (a) diamagnetic (b) paramagnetic (c) ferromagnetic (d) a permanent magnet
8. In crossed fields E = 1 × 10⁴ V m⁻¹ and B = 0.02 T, an electron passes undeflected at a speed of: (a) 2 × 10² m s⁻¹ (b) 5 × 10⁵ m s⁻¹ (c) 2 × 10⁶ m s⁻¹ (d) 5 × 10³ m s⁻¹

Answers

1. (a) B ∝ I/R, and both double.
2. (c) r ∝ v.
3. (b) T = 2πm/(qB) has no v in it.
4. (d) The force is always perpendicular to the velocity.
5. (b) B = μ₀nI, and nI is unchanged.
6. (c) Plane parallel to B means θ = 90° between the normal and B: τ = 50 × 2 × 0.01 × 0.2 = 0.2 N m.
7. (a) Diamagnetic materials have small negative χ.
8. (b) v = E/B = 10⁴/0.02 = 5 × 10⁵ m s⁻¹.

What to do next

• Write the field table (wire, loop, arc, solenoid) and derive the arc result for a semicircle and a quarter circle.
• Solve five radius-and-period questions comparing electrons, protons, deuterons and alpha particles.
• Practise ammeter and voltmeter conversion until you can write both formulas from the circuit picture.
• Learn the magnetic materials table with one example of each, then move on to electromagnetic induction and AC.

For the circuit rules used in galvanometer problems, see current electricity for NEET.