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3 years ago

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A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that a

n 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

1 answer:

Reika [66]3 years ago

3 0

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

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Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

Please hit that thank you button if that helped, I don’t want thank you’s I just want to know that this helped.

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