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 0.43 V and a current of 3.1 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 (a) emf and (b) current are induced in the square coil

Answer 1

Answer:

2.62A

Explanation:

Given

V = 0.43 V

I = 3.1 A

Then, V = IR, R = V/I

R = 0.43/3.1

R = 0.14 Ω

The induced emf = dB/dt * A

So that, dB/dt = emf/A

Since dB/dt is constant then Emf/A(circle) = Emf/A square

So Emf (square)/Emf (circle) = A square / A circle

A circle = πr². The perimeter of the square is 2πr which also is the circumference of the square.

Since the perimeter is 2πr, then each side would be πr/2. Thus, the area of the square would be, (πr/2)² = π²r²/4

So A square/Acircle = (π²r²/4) / πr² = π/4 = 0.79

this means that, emf square = emf circle * 0.79

emf square = 0.43*0.79 = 0.34V

I = V/R

I = 0.34/0.13

I = 2.62A