A magnetic field is perpendicular to the plane of a single-turn circular coil. T

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 0.27 V and a current of 2.1 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 a What (a) emf and (b) current are induced in the square coil? (again perpendicular to the plane of the coil and with magnitude changing at the same rate). (a) Number (b) Number Click if you would like to Show Work for this question: Units Units Open Show Work

Explanation / Answer

Given,

emf = 0.27 V ; I = 2.1 A ;

a)The resistance of the coil will be:

R = emg/I = 0.27/2.1 = 0.13 Ohm

We know that,

emf = d(phi)/dt = A dB/dt

dB/dt = emf/A

emf(sq)/emf(coil) = A(sq)/A(coil)

emf(sq) = emf(coil)[A(sq)/A(coil)]

A(sq) = side^2 ; A(coil) = pi r^2

2 pi r = 4 x side => side = pi r/2

A(sq)/A(coil) = (pi r/2)^2/(pi r^2) = pi/4 = 3.14/4 = 0.785

emf(sq) = emf(coil)[A(sq)/A(coil)]

emf(sq) = 0.27 x 0.785 = 0.212 V

I = emf(sq)/R = 0.212/0.13 = 1.63 A

Hence, emf = 0.212 V ; I = 1.63 A

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