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 0.63 V and a current of 2.7 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 and current are induced in the square coil?

Question

help is greatly appreciated!

Answer 1

ε = 0.63V. I = 2.7 A.
ε = -dÔ/dt =-d(B•S•cosα)/dt.
α= 0 => cos α =1,
ε = - S •dB/dt.
If S is the area of the circle,
then S = π •r², => r=sqrt(S/π).
Perimeter (the length of wire) is
L = 2 •π•r = 2 •π• sqrt(S/π).
the side of the square is
L/4 =2 •π• sqrt(S/π)/4 = π• sqrt(S/π)/2
The area of the square is
S1 = {π• sqrt(S/π)/2}² = π²•S/ 4•π = =π•S/4.
ε1= - S1 •dB/dt= -(π•S/4)• dB/dt,
ε1/ε = [(π•S/4) • (dB/dt)] / {S •dB/dt}= =π/4.
ε1 = ε• π/4 = 0.63• π/4 = 0.495 V.

ε = I•R, => R = ε/I = 0.63/2.7 = 0.23 Ω,
ε1 = I1•R,
I1 = ε1/R = 0.495/0.23 = 2.12 A.