Question

A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.74 V and a current of 3.2 A are induced in the coil. The wire is then re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf is induced in the square coil?

Answer 1

To solve this problem we will apply the concepts related to magnetic flux and induced voltage. This last expression understood as the variation of the magnetic flux over time and, in turn, the magnetic flux expressed as the variation of the magnetic field in a certain area.

Magnetic flux through the circular coil is given as

$\Phi_C = B(\pi r^2)$

The induced voltage is the change of the magnetic flux across the time, then

$\epsilon_{emf,C} = \frac{B(\pi r^2)}{t}$

At the same time the magnetic flux through the square coil would be given as,

$\Phi_S = B(r^2)$

And the induced voltage EMF will be

$\epsilon_{emf,s} = \frac{B(r^2)}{t}$

Equating both expression we have

$\epsilon_{emf,s} = \frac{\epsilon_{emf,C}tr^2}{\pi r^2t}$

$\epsilon_{emf,s} = \frac{0.74V}{\pi}$

$\epsilon_{emf,s} = 0.23355V$

Therefore the emf induced in the square coil is 0.23355V