Electromagnetism: Class 12 Physics Notes, Short Questions and Numericals

Magnetic field and flux density, force on a moving charge and on a current-carrying wire, torque on a coil, galvanometer, ammeter and voltmeter, Ampere's law, solenoid field, and the Hall effect.

Key points

A moving charge or a current produces a magnetic field around it. The field is a vector and is shown by magnetic field lines. Outside a magnet the lines go from north to south. They form closed loops and never cross. Where lines are closer, the field is stronger. The strength of the field is called magnetic flux density B and its SI unit is the tesla (T).

A charge q moving with velocity v in a magnetic field B feels the force F = qvB sinθ, where θ is the angle between v and B. The force is perpendicular to both v and B, so it does no work and does not change the speed. It only changes the direction. If v is along B, the force is zero. One tesla is the field that exerts 1 N on a charge of 1 C moving at 1 m/s at right angles to the field.

When a charge enters a uniform field at right angles, it moves in a circle. The magnetic force supplies the centripetal force: qvB = mv²/r. So the radius is r = mv/(qB), and the time period T = 2πm/(qB) does not depend on speed. This principle is used in the mass spectrometer and the cyclotron.

A straight wire of length L carrying current I in a field B feels the force F = BIL sinθ. The direction is found by Fleming's left hand rule: first finger gives field, second finger gives current, and thumb gives force (motion). Two long parallel wires with currents in the same direction attract, and with opposite directions repel.

The ampere is defined by the force between two long parallel wires. If equal currents in two wires 1 m apart in vacuum give a force of 2 × 10⁻⁷ N on each metre of length, the current in each is 1 A. The force per unit length is F/L = μ₀I₁I₂/(2πd), where μ₀ = 4π × 10⁻⁷ T m/A.

A rectangular coil of N turns, area A and current I in a field B feels a pair of forces that give a torque τ = NIAB cosα, where α is the angle between the plane of the coil and the field (τ = NIAB sinθ if θ is measured from the normal). The torque is greatest when the plane of the coil is parallel to the field. This is the working of the electric motor and the moving coil galvanometer.

Moving coil galvanometer:

A coil is hung between curved pole pieces of a strong magnet, so the field is radial and the torque is the same for any angle.
The current gives a torque NIAB. The spring gives a restoring torque kθ. At balance, NIAB = kθ, so the deflection is proportional to the current.
A soft iron core increases the field and the sensitivity.
A shunt (small resistance in parallel) makes it an ammeter. A multiplier (large resistance in series) makes it a voltmeter.

For a galvanometer of resistance G and full scale current Ig, the shunt S needed to read a larger current I is S = IgG/(I − Ig). The series resistance R needed to read a voltage V is R = V/Ig − G. An ideal ammeter has very low resistance and an ideal voltmeter has very high resistance.

Ampere's law states that the sum of B·Δl around any closed path equals μ₀ times the current passing through the path. For a long straight wire it gives B = μ₀I/(2πr). At the centre of a circular coil of N turns and radius r, B = μ₀NI/(2r).

A solenoid is a long cylindrical coil. Inside it, far from the ends, the field is nearly uniform and given by B = μ₀nI, where n is the number of turns per unit length. The field outside is very small. With an iron core the field becomes much larger. The direction is found by the right hand grip rule: curl the fingers along the current and the thumb shows the field.

The Hall effect: when a current-carrying conductor or semiconductor is placed in a magnetic field perpendicular to the current, a small voltage appears across the sides at right angles to both. This is the Hall voltage. It arises because the magnetic force pushes the charge carriers to one side until an electric field balances it. It is used to find the sign and density of charge carriers and to measure magnetic fields.

Study tip: Practise the direction rules on paper, because many marks are lost on direction. Write the formula, then substitute with units, and convert cm and mA to SI first. For derivations such as the radius of a charged particle path, write the force balance first and then rearrange.

Definitions and flashcards

Say the meaning first, then open the card to check.

Magnetic flux density

The force on a unit charge moving at unit speed perpendicular to the field, in tesla.

Tesla

The field that exerts 1 N on a wire of length 1 m carrying 1 A at right angles to it.

Solenoid

A long cylindrical coil that gives a nearly uniform field inside.

Shunt

A low resistance in parallel with a galvanometer to make an ammeter.

Multiplier

A high resistance in series with a galvanometer to make a voltmeter.

Ampere's law

The line sum of B around a closed path equals μ₀ times the enclosed current.

Hall voltage

The transverse voltage across a current-carrying conductor in a perpendicular field.

Permeability of free space

The constant μ₀ = 4π × 10⁻⁷ T m/A.

Formula sheet

Name	Formula	What it means
Force on a moving charge	F = qvB sinθ	θ is the angle between v and B.
Radius of path	r = mv / (qB)	Charge moving perpendicular to a uniform field.
Force on a wire	F = BIL sinθ	L is the length of wire in the field.
Torque on a coil	τ = NIAB cosα	α is the angle between coil plane and field.
Long straight wire	B = μ₀I / (2πr)	r is the perpendicular distance.
Circular coil centre	B = μ₀NI / (2r)	N turns of radius r.
Solenoid	B = μ₀nI	n is turns per metre.
Shunt	S = IgG / (I − Ig)	To extend the range of a galvanometer to current I.
Multiplier	R = V / Ig − G	To make a voltmeter of range V.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Why does a magnetic force do no work on a moving charge? (2 marks)

The force F = qvB sinθ is always perpendicular to the velocity. The displacement in a short time is along v, so the work F·d is zero. The speed and kinetic energy stay constant, only the direction changes.

Define the tesla. (2 marks)

One tesla is the magnetic flux density that gives a force of 1 newton on a straight wire 1 metre long carrying a current of 1 ampere at right angles to the field.

Why is a radial field used in a moving coil galvanometer? (3 marks)

The pole pieces are curved and a soft iron core is fixed inside the coil. This makes the field radial, so the plane of the coil is always parallel to the field. The torque NIAB is therefore the same at every deflection, and the deflection is directly proportional to the current, which gives a uniform scale.

How is a galvanometer changed into an ammeter and into a voltmeter? (3 marks)

For an ammeter a small resistance, called a shunt, is joined in parallel. Most of the current passes through the shunt, so the instrument has a low total resistance. For a voltmeter a large resistance is joined in series. This limits the current to the full scale value and gives a high total resistance.

Describe the force between two parallel current-carrying wires. (3 marks)

Each wire sets up a field at the other wire, and the other wire then feels a force. Currents in the same direction attract each other and opposite currents repel. The force per unit length is μ₀I₁I₂/(2πd). This relation is used to define the ampere.

Derive the radius of the circular path of a charge in a uniform magnetic field. (5 marks)

Take a charge q of mass m moving with speed v perpendicular to a field B. The magnetic force qvB is always perpendicular to the velocity, so it acts as a centripetal force. Equating, qvB = mv²/r. Cancelling one v and rearranging gives r = mv/(qB). The time for one revolution is T = 2πr/v = 2πm/(qB), which is independent of the speed. A faster particle moves on a larger circle but takes the same time.

Numericals with worked steps

A wire 0.50 m long carries 4.0 A at 30° to a field of 0.20 T. Find the force on it.

F = BIL sinθ.
F = 0.20 × 4.0 × 0.50 × sin 30°.
0.20 × 4.0 × 0.50 = 0.40. sin 30° = 0.5.
F = 0.40 × 0.5 = 0.20 N.

Answer: F = 0.20 N

A proton (mass 1.67 × 10⁻²⁷ kg, charge 1.6 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m/s at right angles to a field of 0.50 T. Find the radius of its path.

r = mv/(qB).
Numerator = 1.67 × 10⁻²⁷ × 2.0 × 10⁶ = 3.34 × 10⁻²¹.
Denominator = 1.6 × 10⁻¹⁹ × 0.50 = 8.0 × 10⁻²⁰.
r = 3.34 × 10⁻²¹ / 8.0 × 10⁻²⁰ = 0.04175 m.

Answer: r ≈ 4.2 × 10⁻² m (about 4.2 cm)

A solenoid has 800 turns on a length of 0.40 m and carries 2.5 A. Find the field inside (μ₀ = 4π × 10⁻⁷ T m/A).

n = 800 / 0.40 = 2000 turns per metre.
B = μ₀nI = 4π × 10⁻⁷ × 2000 × 2.5.
2000 × 2.5 = 5000. 4π × 10⁻⁷ = 1.2566 × 10⁻⁶.
B = 1.2566 × 10⁻⁶ × 5000 = 6.28 × 10⁻³ T.

Answer: B ≈ 6.3 × 10⁻³ T

A galvanometer of resistance 50 Ω gives full scale at 2.0 mA. What shunt makes it read up to 1.0 A?

Ig = 0.002 A, I = 1.0 A, G = 50 Ω.
S = IgG/(I − Ig) = (0.002 × 50) / (1.0 − 0.002).
Numerator = 0.10. Denominator = 0.998.
S = 0.10 / 0.998 = 0.1002 Ω.

Answer: S ≈ 0.10 Ω in parallel

Common mistakes

Using F = qvB when the charge is not at right angles. Include sinθ, and remember the force is zero if v is along B.
Thinking the magnetic force changes the speed of a charge. It changes only the direction.
Connecting a shunt in series or a multiplier in parallel. Shunt is parallel, multiplier is series.
Mixing up Fleming's left hand rule (motors, force) with the right hand rule (generators, induced current).
Using the wrong angle in the coil torque. State clearly whether the angle is from the field to the plane or to the normal.

Practise this chapter

Practice questions and timed test Printable worksheet