Magnetic flux and flux linkageAQA A-Level Physics: Revision notes
Section 1
Magnetic flux
The magnetic flux through a surface is the product of the flux density and the area, when B is normal (perpendicular) to the area:
Φ = BA
The unit is the weber (Wb), where 1 Wb = 1 T m². Flux is a measure of how much field passes through the area. For a uniform field, a large area and a strong field both give large flux.
Section 2
Flux linkage
A coil of N turns links the flux through each turn, so the flux linkage is
NΦ = BAN
measured in Wb turns (or just Wb). It is the quantity that determines the induced emf. For a coil of radius r in a field normal to it, A = πr².
The flux through one turn is Φ = BA. Do not include N until you are asked for the flux linkage.
Section 3
Rotating a coil: NΦ = BAN cos θ
If the normal to the coil makes an angle θ with the field, only the component of B along the normal passes through the area, B cos θ. So
NΦ = BAN cos θ
- θ = 0° (plane perpendicular to field): maximum flux linkage, BAN.
- θ = 90° (plane parallel to field): zero flux linkage.
- For a coil rotating at angular speed ω starting with plane perpendicular to the field, θ = ωt, so NΦ = BAN cos ωt, with ω = 2πf.
The angle in the formula is between the field and the normal to the plane, not between the field and the plane. A plane at 30° to the field has θ = 60°.
Section 4
Worked example
A rectangular coil, 4.0 cm × 6.0 cm, 50 turns, is in a 0.30 T field with its normal at 40° to the field.
A = 2.4 × 10⁻³ m², so NΦ = BAN cos θ = 0.30 × 2.4 × 10⁻³ × 50 × cos 40° = 0.028 Wb turns.
With rotation, use radians on your calculator if you use θ = ωt.
Section 5
Required practical 11: search coil and oscilloscope
Place a small search coil in a uniform alternating magnetic field (for example inside a long solenoid supplied by a signal generator). A changing flux linkage induces an emf in the search coil, which is displayed on an oscilloscope.
- Keep the frequency and the solenoid current constant.
- Rotate the search coil in measured steps of angle θ between its axis and the field.
- Read the peak-to-peak height of the trace (Y-gain × divisions) at each angle.
- Plot peak-to-peak height against cos θ: a straight line through the origin shows the peak flux linkage is proportional to cos θ.
The height is zero (or near zero) at θ = 90°. Repeat readings and use a protractor carefully to reduce uncertainty.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Magnetic flux and flux linkage
- A flat circular coil of 200 turns and radius 3.0 cm is placed in a uniform magnetic field of flux density 0.040 T, with the plane of the coil perpendicular to the field direction.The coil is turned so that the normal to its plane makes an angle of 60° with the field. Calculate the flux linkage.2 marks
- A student uses a small search coil connected to an oscilloscope to investigate how the magnetic flux linkage depends on orientation. The search coil is placed inside a long solenoid carrying a low-frequency alternating current from a signal generator. The search coil can be rotated so that the angle θ between its axis (the normal to its plane) and the field direction can be set to chosen values. For a fixed frequency, the peak-to-peak height of the oscilloscope trace is taken to be proportional to the peak flux linkage.With θ = 0° the peak-to-peak height is 48 mm. Predict the height when θ = 40°.2 marks
- A rectangular coil of 150 turns measures 8.0 cm by 5.0 cm. It is placed in a uniform magnetic field of flux density 0.12 T.Calculate the flux linkage when the plane of the coil is perpendicular to the field, and when the plane of the coil makes an angle of 30° with the field direction.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).