Magnetic Fields

Magnetism and electricity are deeply connected — they are two aspects of the single force of electromagnetism. Moving charges create magnetic fields, and changing magnetic fields can induce electrical currents. This topic covers the motor effect, the force on moving charges, magnetic field patterns, electromagnetic induction, AC electricity, and transformers. It is tested in AQA Paper 2 (Section 7) and OCR Paper 2 (Module 6), and is one of the most mathematically demanding sections of the A-Level course.

Magnetic Flux Density

Key Definition: The magnetic flux density (B) at a point is a measure of the strength and direction of the magnetic field at that point. It is defined by the force exerted on a current-carrying conductor placed in the field.

F = BIl sin θ

where B is the magnetic flux density (measured in tesla, T), I is the current (A), l is the length of conductor in the field (m), and θ is the angle between the current direction and the magnetic field direction. When the conductor is perpendicular to the field (θ = 90°), the force is at its maximum: F = BIl.

One tesla is defined as the magnetic flux density that produces a force of 1 N on a 1 m conductor carrying a current of 1 A at right angles to the field.

Fleming's Left-Hand Rule (Motor Effect)

The direction of the force on a current-carrying conductor in a magnetic field is given by Fleming's left-hand rule:

First finger — direction of the magnetic Field (north to south)
Second finger — direction of the conventional Current (positive to negative)
Thumb — direction of the Thrust (force/motion)

The three fingers are held mutually perpendicular. This rule gives the direction of the force — the conductor is pushed sideways, perpendicular to both the current and the field.

The DC Motor

A simple DC motor uses the motor effect to produce rotation:

A rectangular coil carrying current is placed in a magnetic field.
The two sides of the coil experience forces in opposite directions (by Fleming's left-hand rule), creating a turning moment (torque).
A split-ring commutator reverses the current direction every half turn, ensuring the torque always acts in the same rotational direction.
The coil rotates continuously.

The torque can be increased by: increasing the current, increasing the number of turns, increasing the magnetic field strength, or increasing the area of the coil.

Magnetic Field Patterns

Understanding the shape of magnetic fields around different current-carrying conductors is essential.

Around a long straight wire:

The field lines form concentric circles centred on the wire.
The circles are closer together near the wire (stronger field) and further apart at greater distances (weaker field).
The direction is given by the right-hand grip rule: grip the wire with the right hand so that the thumb points in the direction of conventional current — the fingers curl in the direction of the field lines.
The magnetic flux density at distance r from a long straight wire carrying current I is:

B = μ₀I/(2πr)

where μ₀ = 4π × 10⁻⁷ T m A⁻¹ is the permeability of free space.

Described diagram — Magnetic field pattern around a long straight wire: The wire is shown as a dot (current coming out of the page, marked ⊙) or a cross (current going into the page, marked ⊗) at the centre. Concentric circles surround the wire, with arrowheads showing the direction of the field (anticlockwise for current out of the page, clockwise for current into the page, as given by the right-hand grip rule). The circles are closely spaced near the wire and become more widely spaced further away, indicating that the field strength decreases with distance (B ∝ 1/r).

Inside a solenoid (long coil):

The field inside is uniform — the field lines are straight, parallel, and equally spaced.
The field lines emerge from the north end and loop around to enter the south end (like a bar magnet).
Outside the solenoid, the field is much weaker and non-uniform.
The right-hand grip rule can be adapted: curl the fingers in the direction of current flow, and the thumb points to the north end.
The magnetic flux density inside a solenoid is:

B = μ₀nI

where n is the number of turns per unit length (turns per metre) and I is the current.

Around a flat circular coil:

At the centre, the field is perpendicular to the plane of the coil.
The field pattern resembles that of a short bar magnet from a distance.
Close to the coil, the field lines curve around from one face to the other.

Described diagram — Magnetic field inside a solenoid: A cross-section of the solenoid is shown as a series of circular coil turns viewed from the side. Inside the solenoid, straight parallel horizontal arrows point from left (south end) to right (north end), all equally spaced, indicating a uniform field. Outside the solenoid, the field lines curve around from the north end back to the south end, resembling the external field of a bar magnet. The field is much stronger and uniform inside than outside. The north end is identified using the right-hand grip rule: if the fingers curl in the direction of current flow, the thumb points to the north end.

Exam Tip: Learn the right-hand grip rule thoroughly — it is used to determine the direction of the magnetic field around a wire and to find the polarity (north/south) of a solenoid. Examiners often set questions where you must combine this rule with Fleming's left-hand rule.

Force on a Moving Charge

A charged particle moving through a magnetic field experiences a force:

F = BQv sin θ

where Q is the charge (C) and v is the velocity (m s⁻¹). The force is always perpendicular to both the velocity and the field, so it does no work on the charge (the speed remains constant, but the direction changes).

When a charged particle enters a uniform magnetic field perpendicular to its velocity (θ = 90°), the magnetic force provides the centripetal force, causing the particle to move in a circular path:

BQv = mv²/r

This gives the radius of the circular motion:

r = mv/(BQ)

Key points:

More massive particles have larger orbits.
Faster particles have larger orbits.
Stronger fields produce tighter (smaller) orbits.
The charge-to-mass ratio (Q/m) can be determined by measuring the radius.

This principle is used in cyclotrons (particle accelerators), mass spectrometers (separating isotopes), and the velocity selector (crossed electric and magnetic fields).

If the particle enters at an angle to the field that is neither 0° nor 90°, it follows a helical (spiral) path — circular motion in the plane perpendicular to B combined with constant velocity parallel to B.

Magnetic Flux and Flux Linkage

Key Definition: Magnetic flux (Φ) through a surface is the product of the magnetic flux density and the area perpendicular to the field.

Φ = BA cos θ

where A is the area (m²) and θ is the angle between the magnetic field and the normal to the surface. When the field is perpendicular to the surface (θ = 0°), Φ = BA (maximum). When the field is parallel to the surface (θ = 90°), Φ = 0.

Magnetic flux is measured in webers (Wb), where 1 Wb = 1 T m².

Flux linkage for a coil of N turns is:

NΦ = BAN cos θ

Flux linkage is measured in Wb turns (or simply Wb, since N is dimensionless).

Electromagnetic Induction — Faraday's Law

Key Definition: Faraday's Law states that the magnitude of the induced EMF is equal to the rate of change of magnetic flux linkage through the circuit.

ε = −d(NΦ)/dt

The negative sign reflects Lenz's law (see below).

Methods of Inducing an EMF

An EMF can be induced by:

Moving a conductor through a magnetic field (e.g. pushing a wire through a field)
Changing the magnetic field strength (e.g. moving a magnet towards a coil)
Rotating a coil in a magnetic field (as in a generator)
Changing the area of the coil within the field

flowchart TD
    A["Change in magnetic<br/>flux linkage (NΦ)"] --> B["EMF induced<br/>(Faraday's Law:<br/>EMF = -d(NΦ)/dt)"]
    B --> C["Induced current flows<br/>(if circuit is complete)"]
    C --> D["Current opposes the<br/>change that caused it<br/>(Lenz's Law)"]
    D --> E["Energy is conserved:<br/>work must be done to<br/>maintain the change"]

Worked Example — Induced EMF Calculation

Question: A coil of 200 turns and area 5.0 × 10⁻³ m² is placed perpendicular to a uniform magnetic field that increases uniformly from 0.10 T to 0.50 T in 0.20 s. Calculate the magnitude of the induced EMF.

Solution:

Step 1: Calculate the change in flux linkage. Initial flux linkage = BAN = 0.10 × 5.0 × 10⁻³ × 200 = 0.10 Wb Final flux linkage = 0.50 × 5.0 × 10⁻³ × 200 = 0.50 Wb Change in flux linkage = 0.50 − 0.10 = 0.40 Wb

Step 2: Apply Faraday's law. |ε| = Δ(NΦ)/Δt = 0.40/0.20 = 2.0 V

Lenz's Law

Key Definition: Lenz's Law states that the direction of the induced current (or EMF) is always such that it opposes the change in flux that produces it.

This is a consequence of the conservation of energy. If the induced current aided the change, it would increase the flux change, which would induce more current, creating energy from nothing — violating the First Law of Thermodynamics.

Example: When a north pole of a magnet is pushed towards a solenoid, the induced current flows in a direction that makes the near end of the solenoid a north pole, repelling the approaching magnet. You must do work against this repulsion to push the magnet in, and this work is what provides the electrical energy.

The AC Generator

An AC generator (alternator) consists of a coil rotating in a uniform magnetic field. The flux linkage through the coil varies sinusoidally:

NΦ = BAN cos(ωt)

By Faraday's law, the induced EMF is:

ε = BANω sin(ωt)

The peak EMF is ε₀ = BANω. The output is an alternating voltage that varies sinusoidally with time. The EMF is maximum when the coil is parallel to the field (flux linkage is changing most rapidly) and zero when perpendicular to the field (flux linkage is at a maximum or minimum and momentarily not changing).

Described diagram — AC generator output waveform: The horizontal axis is time (t) and the vertical axis is EMF (ε). The curve is a smooth sinusoidal wave oscillating symmetrically above and below the time axis. It starts at ε = 0 when t = 0 (coil perpendicular to the field), rises to a positive peak of +ε₀ at one quarter of a cycle, returns to zero at half a cycle, falls to a negative peak of −ε₀ at three quarters of a cycle, and returns to zero after one complete cycle. The pattern repeats continuously. The period T = 2π/ω is the time for one full cycle, and the frequency f = 1/T. Increasing the angular speed ω increases both the frequency and the peak EMF.

AC Electricity: Peak and RMS Values

Mains electricity in the UK is alternating current (AC) at a frequency of 50 Hz and a quoted voltage of 230 V. The quoted values for AC are root mean square (rms) values, not peak values.

Key Definition: The rms value of an alternating current or voltage is the value of the steady direct current or voltage that would dissipate energy at the same rate in a resistor.

The relationships between peak (₀) and rms values are:

V_rms = V₀/√2 ≈ 0.707 V₀

I_rms = I₀/√2 ≈ 0.707 I₀

Conversely: V₀ = V_rms × √2, and I₀ = I_rms × √2.

For UK mains: V_rms = 230 V, so V₀ = 230 × √2 ≈ 325 V. The peak-to-peak voltage is about 650 V.

Power in AC Circuits

The average power dissipated in a resistor carrying AC is:

P = I_rms × V_rms = I_rms² × R = V_rms²/R

These formulae have exactly the same form as for DC, provided you use rms values.

Oscilloscope Traces

An oscilloscope displays how voltage varies with time:

DC signal: A horizontal straight line. The height above the zero line gives the voltage.
AC signal: A sinusoidal wave. The peak voltage is measured from the zero line to the crest. The period T is the time for one complete cycle, and the frequency f = 1/T.
From the trace, you can determine: peak voltage (from the amplitude), frequency (from the period), and hence calculate the rms voltage.

The timebase setting (e.g. 5 ms/div) tells you the time per horizontal division. The Y-gain setting (e.g. 2 V/div) tells you the voltage per vertical division.

Measurement	How to Read from Oscilloscope
Peak voltage	Count vertical divisions from zero to peak × Y-gain
Peak-to-peak voltage	Count vertical divisions from trough to peak × Y-gain
Period (T)	Count horizontal divisions for one full cycle × timebase
Frequency (f)	f = 1/T
rms voltage	V₀/√2

Exam Tip: When reading oscilloscope traces, always state the settings you are using (timebase and Y-gain) and show your working. A common error is to measure peak-to-peak voltage and forget to halve it to find the peak voltage.

Transformers

A transformer consists of two coils (primary and secondary) wound around a shared soft iron core. It works on the principle of electromagnetic induction: an alternating current in the primary coil creates a changing magnetic flux in the core, which induces an EMF in the secondary coil.

Key Definition: Transformers only work with AC (not DC), because only a changing current produces a changing magnetic flux needed for induction.

The turns ratio relates the voltages:

V_s/V_p = N_s/N_p

For an ideal (100% efficient) transformer, power in = power out:

V_p I_p = V_s I_s

Step-up transformer: N_s > N_p, so V_s > V_p (voltage is increased, current is decreased)
Step-down transformer: N_s < N_p, so V_s < V_p (voltage is decreased, current is increased)

Worked Example — Transformer Ratio Calculation

Question: A transformer is used to step down the mains voltage of 230 V to 12 V for a model railway. The primary coil has 4600 turns. (a) Calculate the number of turns on the secondary coil. (b) If the model railway draws a current of 2.5 A, calculate the current in the primary coil, assuming the transformer is ideal.

Solution:

(a) Using V_s/V_p = N_s/N_p: N_s = N_p × V_s/V_p = 4600 × 12/230 = 4600 × 0.05217 = 240 turns

(b) Using V_p I_p = V_s I_s: I_p = V_s I_s / V_p = 12 × 2.5 / 230 = 30/230 = 0.13 A

Note that the current is stepped up in the primary (smaller) relative to the secondary in a step-down transformer. Power is conserved: P = 230 × 0.13 = 30 W = 12 × 2.5.

The National Grid

The National Grid transmits electrical power at very high voltage (up to 400 kV in the UK) to reduce current and therefore reduce energy losses due to heating in the cables (P_loss = I²R). Step-up transformers increase the voltage at the power station, and step-down transformers decrease it to safe levels for domestic use (230 V in the UK).

flowchart LR
    A["Power Station<br/>~25 kV"] --> B["Step-Up Transformer<br/>(increases V, decreases I)"]
    B --> C["National Grid<br/>Transmission Lines<br/>~400 kV<br/>(low I = low I²R losses)"]
    C --> D["Step-Down Transformer<br/>(decreases V, increases I)"]
    D --> E["Homes and Businesses<br/>~230 V"]

If voltage is doubled, current is halved (for the same power), and I²R losses decrease by a factor of four. This is why high-voltage transmission is essential for efficient power distribution.

Sources of Energy Loss in Transformers

Source of Loss	Cause	How to Reduce
Eddy currents	Induced currents in the iron core generate heat	Laminate the core (layers separated by insulation)
Resistance of windings	Current flowing through the copper coils generates heat (I²R)	Use thick, low-resistance copper wire
Hysteresis losses	Energy is needed to repeatedly magnetise and demagnetise the core	Use soft iron (easily magnetised and demagnetised)
Flux leakage	Not all the magnetic flux passes through both coils	Wind coils tightly on the same core; use a toroidal core

Exam Tip: Real transformers are typically 95–99% efficient. In calculations, if you are told the transformer is "ideal," use 100% efficiency. If given an efficiency, multiply the ideal secondary power by the efficiency: P_out = η × P_in, where η is the efficiency expressed as a decimal.