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 0.65 V and a current of 3.8 A are induced in the coil. The wire is the 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 and current are induced in the square coil?

Answer 1

Answer:

$E_{square}=0.51v$

$i_{square}=2.98A$

Explanation:

$E_{mf}=-N*\frac{d\alpha}{dt}$

$E_{mf}=-N*A*\frac{d\beta}{dt}$

$N=1$

$E_{circle}=A_{circle}*\frac{d\beta}{dt}$

$E_{square}=E_{circle}*\frac{A_{square}}{A_{circle}}$

$E_{square}=E_{circle}(\frac{\frac{\pi^2}{4}*r^2}{\pi*r^2})$

$E_{square}=E_{circle}*\frac{\pi}{4}$

$E_{square}=0.65v*\frac{\pi}{4}$

$E_{square}=0.51v$

$i_{square}=i_{circle}*\frac{E_{square}}{E_{circle}}$

$i_{square}=3.8A*\frac{0.51v}{0.65v}$

$i_{square}=2.98A$