Moving Charges and Magnetism — Class 12 Physics Worksheet | AAI ATC

Before you begin

Formulas you will need

Key formulas B = μ₀I ∕ 2πr (long wire) · B = μ₀NI ∕ 2R (centre of coil) · B = μ₀nI (solenoid)
F = qvB sinθ · F = BIL sinθ · τ = NIAB sinθ · m = NIA
Radius of circular path: r = mv ∕ qB · Force per length between parallel wires: μ₀I₁I₂ ∕ 2πd
μ₀ = 4π × 10⁻⁷ T m A⁻¹

How to use this worksheet. Torque uses the angle between the normal to the coil and the field, not the plane of the coil. And when a wire crosses a circular field region off-centre, the length inside the field is a chord, not the diameter — that single idea is the whole of the first previous year question.

Section A · Level 1

Definitions and direct formulas

Ten questions. None should take more than twenty seconds.

Q1

The SI unit of magnetic field strength is the:

Magnetic flux density is measured in tesla, which is one weber per square metre. The gauss is a smaller non-SI unit equal to 10⁻⁴ T.

Q2

The magnetic force on a charge q moving with velocity v at an angle θ to a field B is:

The force is the cross product of velocity and field, so its magnitude is qvB sinθ. It is greatest when the motion is perpendicular to the field.

Q3

The magnetic force on a stationary charge placed in a magnetic field is:

Since the force depends on velocity, a charge at rest experiences no magnetic force at all. This is the clearest difference between magnetic and electric forces.

Q4

The Biot-Savart law gives the:

The Biot-Savart law expresses the magnetic field contributed by a small current-carrying element, and integrating it gives the field of a whole conductor.

Q5

The magnetic field at the centre of a circular coil of N turns, radius R, carrying current I is:

At the centre of a circular coil the field is μ₀NI divided by 2R. Option 4 is the field of a long straight wire, which is the standard distractor.

Q6

The magnetic field inside a long solenoid with n turns per unit length carrying current I is:

Inside a long solenoid the field is uniform and equal to μ₀nI, independent of the radius and of the position along the axis away from the ends.

Q7

Two long parallel wires carrying current in the same direction will:

Parallel currents in the same direction attract; antiparallel currents repel. This is the effect used to define the ampere.

Q8

The work done by a magnetic force on a moving charged particle is:

The magnetic force is always perpendicular to the velocity, so it can change the direction of motion but never the speed. No work is done and the kinetic energy stays constant.

Q9

A moving charge travelling parallel to a uniform magnetic field experiences a force of:

With θ = 0 the sine term vanishes, so the force is zero. The particle simply continues in a straight line.

Q10

A galvanometer is converted into an ammeter by connecting a:

An ammeter must have low resistance and carry large currents, so a small shunt resistance is placed in parallel. A voltmeter is the opposite case, needing a high resistance in series.

Section B · Level 2

Fields, forces and torques

Nine questions. Substitute carefully and watch the powers of ten.

Q11

A straight wire of length 0.4 m carrying 2 A lies perpendicular to a field of 0.5 T. The force on it is:

F = BIL = 0.5 × 2 × 0.4 = 0.4 N. The wire is perpendicular to the field, so the sine term is 1.

Q12

A charge of 1.6 × 10⁻¹⁹ C moves at 10⁶ m/s perpendicular to a 0.2 T field. The force on it is:

F = qvB = 1.6 × 10⁻¹⁹ × 10⁶ × 0.2 = 3.2 × 10⁻¹⁴ N. Track the powers of ten carefully.

Q13

The magnetic field at the centre of a single circular loop of radius 0.1 m carrying 2 A is closest to:

B = μ₀I ∕ 2R = (4π × 10⁻⁷ × 2) ∕ 0.2 = 1.26 × 10⁻⁵ T.

Q14

A solenoid has 1000 turns per metre and carries 2 A. The field inside is closest to:

B = μ₀nI = 4π × 10⁻⁷ × 1000 × 2 = 2.51 × 10⁻³ T.

Q15

The radius of the circular path of a charged particle moving perpendicular to a magnetic field is:

Setting the magnetic force equal to the centripetal force, qvB = mv² ∕ r, gives r = mv ∕ qB. Faster or heavier particles curve less.

Q16

Two long parallel wires 1 m apart each carry 1 A. The force per unit length between them is:

Force per unit length is μ₀I₁I₂ ∕ 2πd = (4π × 10⁻⁷ × 1 × 1) ∕ (2π × 1) = 2 × 10⁻⁷ N/m. This value is exactly how the ampere used to be defined.

Q17

The torque on a current loop of N turns, area A, carrying current I in a field B, when the plane of the loop is parallel to the field, is:

Torque is NIAB sinθ where θ is measured between the normal to the loop and the field. If the plane is parallel to B, the normal is perpendicular to B, so θ = 90° and the torque is maximum at NIAB.

Q18

The magnetic dipole moment of a coil of 10 turns and area 0.01 m² carrying 2 A is:

m = NIA = 10 × 2 × 0.01 = 0.2 A m².

Q19

A galvanometer is converted into a voltmeter by connecting a:

A voltmeter must draw almost no current, so a large resistance is placed in series with the galvanometer coil.

Section C · Level 3

Motion of charges and field geometry

Eight questions. Each needs two steps or a physical insight.

Q20

A coil of 100 turns and radius 0.05 m carries 1 A. The field at its centre is closest to:

B = μ₀NI ∕ 2R = (4π × 10⁻⁷ × 100 × 1) ∕ 0.1 = 1.26 × 10⁻³ T. Multiplying by N is the step most often missed.

Q21

The torque on a current loop is maximum when the normal to the loop makes an angle with the field of:

Torque follows sinθ measured from the normal, so it peaks at 90° and is zero at 0° and 180°, where the loop is in equilibrium.

Q22

The frequency of revolution of a charged particle in a cyclotron is:

The cyclotron frequency is qB divided by 2πm. Notice that it does not depend on speed or radius, which is exactly why the cyclotron works.

Q23

A wire passes through a cylindrical field region of radius 10 cm at a perpendicular distance of 6 cm from the axis. The length of wire inside the field is:

The chord length is 2√(R² − d²) = 2√(0.01 − 0.0036) = 2 × 0.08 = 0.16 m. This chord calculation is the heart of one of the previous year questions.

Q24

A proton and an alpha particle enter the same magnetic field with the same speed, perpendicular to the field. Compared with the proton, the alpha particle follows a path of radius:

Radius is mv ∕ qB. The alpha particle has four times the mass and twice the charge, so the ratio m/q doubles and the radius doubles.

Q25

A charged particle moves in a helical path in a uniform magnetic field. This happens when its velocity is:

The component along the field is unaffected while the perpendicular component curves in a circle. Combining the two gives a helix.

Q26

Inside a long current-carrying solenoid, the magnetic field:

The field inside a long solenoid is essentially uniform and equal to μ₀nI everywhere in the cross-section, weakening only near the ends.

Q27

The magnetic field at a perpendicular distance r from a long straight wire carrying current I is proportional to:

For a long straight wire, B = μ₀I ∕ 2πr, so the field falls off as one over r. Confusing this with an inverse square is a common error.

Section D · Previous year questions

Straight from the AAI ATC papers

Three questions, every one exactly as it appeared in the 2021 papers.

Q28PYQ 2021 · Shift 2

A uniform magnetic field of 2 T exists in a cylindrical region of radius 10 cm, its direction parallel to the axis along east to west. A wire carrying current 6 A in the north to south direction passes through the region. What is the magnitude of the force on the wire if (a) the wire intersects the axis, and (b) the wire in the N-S direction is lowered from the axis by a distance of 6 cm?

In case (a) the wire runs along a diameter, so the length in the field is 0.2 m and F = BIL = 2 × 6 × 0.2 = 2.4 N. In case (b) the wire cuts a chord of length 2√(0.1² − 0.06²) = 0.16 m, giving F = 2 × 6 × 0.16 = 1.92 N. Note that lowering the wire shortens the chord but does not change the field strength.

Q29PYQ 2021 · Shift 2

A circular coil of 50 turns and radius 10 cm carrying a current of 5.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 90° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

The area is πr² = π × 0.01 = 0.0314 m². With θ = 90° the sine is 1, so τ = NIAB = 50 × 5 × 0.0314 × 1 = 7.85 N m. The counter torque must match this exactly.

Q30PYQ 2021 · Shift 3

A galvanometer coil has a resistance of 14 Ω and shows full scale deflection for a current of 4 mA. What series resistance converts it into a voltmeter of range 0 to 16 V?

The total resistance needed is 16 ∕ 0.004 = 4000 Ω. The coil already provides 14 Ω, so the series resistance is 4000 − 14 = 3986 Ω. Option 2 forgets to subtract the coil resistance.

↓ Free download — Moving Charges and Magnetism PDF

All thirty questions with the formula sheet printed on the front page and ruled working space for your circuit reductions. Print it, solve it, keep it.

Download the free PDF Join our live classes — extremely affordable