3.7.5.3
Worked example: pulling it all together
Magnetic Flux & Flux Linkage — AQA A-Level Physics
- Exam questions on this spec point typically combine flux, flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns)., and the angle. Here's a multi-part example that ties the whole section together.
Worked Example
A flat circular coil has 150 turns and a radius of 4.0 cm. It is placed in a uniform magnetic field of flux densityMass per unit volume of a material. Measured in kg m⁻³. 0.25 T.
(a) Calculate the maximum flux through one turn of the coil.
(b) Calculate the maximum flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns). for the coil.
(c) The coil is rotated so the normal makes an angle of 60° with the field. Calculate the new flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns)..
Show Solution
1
Calculate the area of the coil
The coil is circular with radius $r = 4.0 \text{ cm} = 0.040 \text{ m}$
$$A = \pi r^2 = \pi \times (0.040)^2 = 5.027 \times 10^{-3} \text{ m}^2$$2
(a) Maximum flux through one turn
Maximum flux occurs when the field is perpendicular to the coil face ($\theta = 0°$):
$$\Phi = BA \cos 0° = BA$$ $$\Phi = 0.25 \times 5.027 \times 10^{-3}$$ $$= 1.3 \times 10^{-3} \text{ Wb (2 s.f.)}$$3
(b) Maximum flux linkage
Multiply by the number of turns:
$$N\Phi = N \times BA = 150 \times 1.257 \times 10^{-3}$$ $$= 0.19 \text{ Wb turns (2 s.f.)}$$4
(c) Flux linkage at θ = 60°
$$N\Phi = BAN \cos \theta$$
$$= 0.25 \times 5.027 \times 10^{-3} \times 150 \times \cos 60°$$
Since $\cos 60° = 0.50$:
$$= 0.1885 \times 0.50$$ $$= 0.094 \text{ Wb turns (2 s.f.)}$$Answer
(a) $\Phi = 1.3 \times 10^{-3}$ Wb (b) $N\Phi = 0.19$ Wb turns (c) $N\Phi = 0.094$ Wb turns
Examiner Tips and Tricks
- Part (c) is exactly half of part (b).
- That's because cos 60° = 0.5.
- Examiners love using 60° and 30° because the trig values are clean.
- If your answer doesn't simplify neatly with these angles, double-check your working.