Question

A 1000-turn coil of wire 2.0 cm in diameter is in a magnetic field that drops from 0.10 T to 0 T in 10 ms. The axis of the coil is parallel to the field. What is the emf of the coil

Answer 1

Emf of the coil is 12.57

The Emf is represented by e.

e= N * dΦ / dt

where N is the number of turns.

Φ= B.A= B * A *cos θ

where θ is the angle between field and area.

area vector is parallel to the axis and also the field is parallel to the axis.

hence, the area vector is also parallel to the field.

then, θ becomes zero.

cos (0)= 1

hence, Φ= B * A * 1= B * A

therefore, the formula for Emf becomes,

e= N * d(B * A) / dt

 = N * A * d(B) / dt     since A is constant

 = 1000 * $\pi$ * $2^{2}$ * $10^{-4}$ * (0.10 - 0 / 10)  * $10^{3}$

 =12.57 V

for better understanding click the link below:

brainly.com/question/26334813

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