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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.
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.
The direction of the force on a current-carrying conductor in a magnetic field is given by Fleming's left-hand rule:
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.
A simple DC motor uses the motor effect to produce rotation:
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.
Understanding the shape of magnetic fields around different current-carrying conductors is essential.
Around a long straight wire:
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):
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:
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.
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:
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.
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).
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).
An EMF can be induced by:
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"]
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
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.
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.
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.
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.
An oscilloscope displays how voltage varies with time:
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.
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
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 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.
| 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.