Magnetic Fields — A-Level Physics Study Guide

For: A-Level Physics candidates sitting A-Level Physics.

Covers: Force on current-carrying conductors, force on moving charges, magnetic flux density, Faraday's and Lenz's laws of electromagnetic induction, AC generation, and transformers as per the A-Level Physics syllabus.

You should already know: IGCSE Physics, basic algebra and trigonometry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Physics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Are Magnetic Fields?

A magnetic field is a region of space where a magnetic force is exerted on magnetic materials, current-carrying conductors, or moving charged particles. Often referred to as a B-field, it is represented by magnetic field lines that point from the north to south pole of a magnet outside the material, and south to north inside the magnet. This topic accounts for 8-10% of total marks across Paper 2 (AS Level), Paper 4 (A Level), and Paper 5 (practical assessment) of the A-Level Physics exam series.

2. Force on a current-carrying conductor — $F=BIL\sin \theta$

When current flows through a conductor placed in an external magnetic field, the moving charges in the current experience a force that is transferred to the conductor as a whole. This is called the motor effect, and its direction is given by Fleming’s Left Hand Rule (FLHR): hold your first finger (magnetic field direction), second finger (conventional current direction), and thumb mutually perpendicular, and the thumb points in the direction of the force.

The magnitude of the force is calculated with the formula: $$F=BIL\sin \theta$$ Where:

• $F$ = magnetic force in newtons (N)
• $B$ = magnetic flux density in tesla (T, defined in Section 4)
• $I$ = current in the conductor in amperes (A)
• $L$ = length of the conductor located inside the magnetic field in meters (m)
• $\theta$ = angle between the direction of the current and the magnetic field lines

Force is maximum when $\theta =90^{\circ }$ ($\sin \theta =1$) and zero when the current is parallel to the field ($\theta =0^{\circ }$, $\sin \theta =0$).

Worked Example: A 50 cm long wire carrying a current of 2.4 A is placed at 30° to a uniform magnetic field of flux density 0.15 T. Calculate the force on the wire.

1. Convert length to SI units: $50\text{cm}=0.5\text{m}$
2. Substitute values: $F=0.15\times 2.4\times 0.5\times \sin (30^{\circ })=0.15\times 2.4\times 0.5\times 0.5=0.09\text{N}$

3. Force on a moving charge — $F=qvB$

This formula is a direct derivation of the force on a current-carrying conductor, since current is defined as $I=\frac{\Delta Q}{\Delta t}$. Substitute this into $F=BIL\sin \theta$, and note that the length of the conductor a charge travels through in time $\Delta t$ is $L=v\Delta t$, where $v$ is the speed of the charge. Cancelling $\Delta t$ gives the force on a single moving charge: $$F=qvB\sin \theta$$ Where:

• $q$ = charge of the particle in coulombs (C)
• $v$ = speed of the particle in m/s
• $\theta$ = angle between the particle’s velocity vector and the magnetic field

Since this force is always perpendicular to both the velocity of the particle and the magnetic field, it acts as a centripetal force, causing charged particles to move in circular paths when moving perpendicular to a uniform B field. Examiners frequently ask for the derivation of the radius of this circular path: equate $qvB=\frac{m{v}^2}{r}$ to get $r=\frac{mv}{qB}$, where $m$ is the mass of the particle.

Worked Example: An electron (charge $1.6\times 10^{-19}\text{C}$) moves at $3.0\times 10^6\text{m/s}$ perpendicular to a 0.2 T magnetic field. Calculate the force on the electron.

1. Since velocity is perpendicular to B, $\sin \theta =1$
2. Substitute values: $F=1.6\times 10^{-19}\times 3.0\times 10^6\times 0.2=9.6\times 10^{-14}\text{N}$

4. Magnetic flux density and Tesla

Magnetic flux density $B$ is the measure of the strength of a magnetic field, defined as the force per unit current per unit length of a conductor placed perpendicular to the field. Its SI unit is the tesla (T), formally defined as:

1 tesla is the magnetic flux density that produces a force of 1 N on a 1 m long conductor carrying a current of 1 A perpendicular to the field. $$1\text{T}=1\text{N}{\text{A}}^{-1}{\text{m}}^{-1}$$

You will also encounter magnetic flux $\Phi$, which is the total magnetic field passing through a given area, calculated as: $$\Phi =BA\cos \theta$$ Where $A$ is the area of the surface, and $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the surface. The unit of flux is the weber (Wb), so $1\text{Wb}=1\text{T}{\text{m}}^2$. For a coil with $N$ turns, the total magnetic flux linkage is $N\Phi$, measured in weber-turns.

Worked Example: A square coil of side 10 cm is placed with its plane perpendicular to a 0.5 T magnetic field. Calculate the magnetic flux through the coil.

1. Calculate area: $A=0.1\text{m}\times 0.1\text{m}=0.01{\text{m}}^2$
2. The normal to the coil is parallel to B, so $\theta =0^{\circ }$, $\cos \theta =1$
3. Substitute: $\Phi =0.5\times 0.01\times 1=5\times 10^{-3}\text{Wb}$

5. Electromagnetic induction — Faraday's and Lenz's laws

Electromagnetic induction is the process of generating an electromotive force (emf) by changing the magnetic flux linked with a conductor or coil, forming the basis of generators and transformers. There are two core laws governing this effect:

1. Faraday’s Law of Electromagnetic Induction: The magnitude of the induced emf is equal to the rate of change of magnetic flux linkage with the circuit. $$\epsilon =\left|\frac{\Delta (N\Phi )}{\Delta t}\right|$$ The modulus sign indicates we are calculating magnitude first, before finding direction with Lenz’s Law.
2. Lenz’s Law: The direction of the induced emf (and induced current, if the circuit is closed) is such that it opposes the change that caused it. This is a consequence of conservation of energy: if the induced current did not oppose the change, you would generate infinite free energy, which is impossible.

Worked Example: A coil of 200 turns has a flux through it that changes from 2 mWb to 10 mWb in 0.4 s. Calculate the magnitude of the induced emf.

1. Convert flux values: $2\text{mWb}=2\times 10^{-3}\text{Wb}$, $10\text{mWb}=1\times 10^{-2}\text{Wb}$
2. Calculate change in flux linkage: $\Delta (N\Phi )=200\times (1\times 10^{-2}-2\times 10^{-3})=1.6\text{Wb turns}$
3. Substitute: $\epsilon =\frac{1.6}{0.4}=4\text{V}$

6. AC generation and transformers

AC Generators

An AC generator (alternator) converts mechanical energy to electrical energy by rotating a coil in a uniform magnetic field, which continuously changes the flux linkage with the coil to induce an alternating emf. The induced emf as a function of time is: $$\epsilon ={\epsilon }_0\sin (\omega t)$$ Where ${\epsilon }_0=NAB\omega$ is the peak emf, $\omega =2\pi f$ is the angular speed of rotation, and $f$ is the frequency of the alternating current.

Transformers

Transformers change the voltage of an alternating current supply using the principle of mutual induction, consisting of two insulated coils (primary and secondary) wound on a laminated soft iron core. For an ideal (100% efficient) transformer:

1. Turns ratio rule: $\frac{V_s}{V_p}=\frac{N_s}{N_p}$, where $V$ is voltage, $N$ is number of turns, and subscripts $s$ and $p$ refer to secondary and primary coils respectively. Step-up transformers have $N_s>N_p$, step-down transformers have $N_s<N_p$.
2. Power conservation rule: $V_p{I}_p=V_s{I}_s$, since no power is lost in ideal transformers.

Real transformers have small energy losses from eddy currents in the core, coil resistance, and flux leakage, which can be reduced by using laminated cores, low-resistance copper wire, and tightly wound coils.

Worked Example 1 (AC Generator): A coil of 50 turns, area 0.02 m², rotates at 50 Hz in a 0.3 T magnetic field. Calculate the peak emf.

1. Calculate angular speed: $\omega =2\pi \times 50=100\pi \text{rad/s}$
2. Substitute: ${\epsilon }_0=50\times 0.02\times 0.3\times 100\pi \approx 94.2\text{V}$

Worked Example 2 (Transformer): A step-down transformer has 2000 turns on primary, 100 turns on secondary, connected to 240 V AC supply. Calculate secondary voltage, and secondary current if primary current is 0.5 A (ideal transformer).

1. Secondary voltage: $V_s=240\times \frac{100}{2000}=12\text{V}$
2. Secondary current: $I_s=0.5\times \frac{2000}{100}=10\text{A}$

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Using Fleming’s Left Hand Rule for induced current in generators instead of the Right Hand Rule. Why students do it: They mix up the motor effect (force produced from current) and generator effect (current produced from motion). Correct move: Use FLHR for motors (current supplied, force output), Fleming’s Right Hand Rule for generators (motion supplied, current output).
• Pitfall 2: Using the angle between the coil plane and B field instead of the angle between the normal to the coil and B field in flux calculations. Why students do it: Exam questions often give the angle of the coil plane, not the normal. Correct move: If the coil plane is at angle $\theta$ to B, the normal is at $90^{\circ }-\theta$, so use $\cos (90^{\circ }-\theta )=\sin \theta$ in $\Phi =BA\cos \theta$.
• Pitfall 3: Applying the transformer turns ratio to DC supplies. Why students do it: They forget transformers rely on changing flux to induce emf. Correct move: DC produces constant flux, so no emf is induced in the secondary coil, transformers cannot step up/down DC voltage.
• Pitfall 4: Forgetting the $\sin \theta$ term in force calculations, assuming $F=BIL$ always applies. Why students do it: Many practice questions use perpendicular current and B field, so students get used to omitting the term. Correct move: Always check the angle between current/velocity and B field, include $\sin \theta$ unless explicitly told the two are perpendicular.
• Pitfall 5: Mixing up units of flux density and flux, writing Wb for B instead of T. Why students do it: The two terms are closely related. Correct move: Remember $B$ = flux density (units T, $N{A}^{-1}{m}^{-1}$), $\Phi$ = flux (units Wb, $T{m}^2$).

8. Practice Questions (A-Level Physics Style)

Question 1 (4 marks)

A copper rod of length 0.75 m is suspended horizontally by two flexible leads in a uniform vertical magnetic field of flux density 0.18 T. The rod has mass 0.04 kg. Calculate the magnitude and direction of the current required in the rod to make the tension in the leads zero. Use $g=9.81{\text{m/s}}^2$.

Solution

Tension is zero when the upward magnetic force equals the downward weight of the rod: $BIL\sin \theta =mg$.

1. Current is horizontal, B is vertical, so $\theta =90^{\circ }$, $\sin \theta =1$
2. Rearrange for current: $I=\frac{mg}{BL}=\frac{0.04\times 9.81}{0.18\times 0.75}\approx 2.91\text{A}$
3. Direction: Using Fleming’s Left Hand Rule, with force upward and B field downward, current flows horizontally left to right (or perpendicular to both B and force, consistent with FLHR).

Question 2 (5 marks)

A circular coil of 120 turns, radius 0.05 m, is placed with its plane perpendicular to a uniform magnetic field of flux density 0.25 T. The field is reduced uniformly to zero in 0.12 s. a) Calculate the magnitude of the induced emf. b) State Lenz’s Law and use it to explain the direction of the induced current.

Solution

a) 1. Calculate coil area: $A=\pi r^2=\pi \times (0.05{)}^2\approx 7.85\times 10^{-3}{\text{m}}^2$ 2. Initial flux linkage: $N{\Phi }_i=120\times 0.25\times 7.85\times 10^{-3}\approx 0.236\text{Wb turns}$ 3. Induced emf: $\epsilon =\frac{\Delta (N\Phi )}{\Delta t}=\frac{0.236}{0.12}\approx 1.97\text{V}$

b) Lenz’s Law states that the direction of the induced current opposes the change that caused it. The change here is a decreasing magnetic field through the coil, so the induced current will produce a magnetic field in the same direction as the original field to oppose the decrease, with direction found using the right-hand grip rule for coils.

Question 3 (4 marks)

An ideal transformer is used to step down 11 kV AC to 230 V AC for domestic supply. The maximum power supplied to households is 230 kW. Calculate: a) The turns ratio of primary to secondary b) The maximum current in the secondary coil c) The maximum current in the primary coil

Solution

a) Turns ratio: $\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{11000}{230}\approx 47.8:1$ b) Secondary current: $I_s=\frac{P}{V_s}=\frac{230000}{230}=1000\text{A}$ c) For ideal transformer, input power = output power, so $I_p=\frac{P}{V_p}=\frac{230000}{11000}\approx 20.9\text{A}$

9. Quick Reference Cheatsheet

Units

• $1\text{T}=1\text{N}{\text{A}}^{-1}{\text{m}}^{-1}$
• $1\text{Wb}=1\text{T}{\text{m}}^2$

10. What's Next

Magnetic fields is a foundational topic that connects directly to several later sections of the A-Level Physics syllabus, including charged particle physics (where you will use the circular motion of charged particles in B fields to analyze particle accelerators and mass spectrometers), electromagnetism in practical circuits, and medical physics applications such as MRI scanners. You will also encounter these concepts in Paper 5 practical assessments, where you may be asked to design experiments to measure magnetic flux density or transformer efficiency.

To reinforce your understanding, you can practise past A-Level Physics questions on magnetic fields from Paper 2 and Paper 4, paying close attention to mark scheme conventions for explaining Lenz’s Law and transformer losses. If you have any questions about concepts, worked examples, or exam technique, you can ask Ollie anytime on the homepage, where you can also access more topic guides, practice quizzes, and past paper walkthroughs tailored to A-Level Physics.

Aligned with the Cambridge International AS & A Level Physics 9702 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.