roblem 21.11 A flat, square coil with 17.0 turns has sides of length 0.140 m . T

Question

roblem 21.11

A flat, square coil with 17.0 turns has sides of length 0.140 m . The coil rotates in a magnetic field of 2.00×102 T .

Part A

What is the angular velocity of the coil if the maximum emf produced is 20.0 mV ?

3.00

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Correct

Part B

What is the average emf at the angular velocity found in part A?

please help with part B

roblem 21.11

A flat, square coil with 17.0 turns has sides of length 0.140 m . The coil rotates in a magnetic field of 2.00×102 T .

Part A

What is the angular velocity of the coil if the maximum emf produced is 20.0 mV ?

3.00

SubmitHintsMy AnswersGive UpReview Part

Correct

Part B

What is the average emf at the angular velocity found in part A?

please help with part B

Explanation / Answer

emf = - rate of change of flux = - rate of change of (B x Area x costheta)

theta = omega x t

So, Differentiate and get

emf = B x A x omega x sin(omega t) which has a max when the sine = 1

so 20.0 * 10-3 = 2.0 * 10-2 * (0.140)2 * 17 *omega

solve for omega (3 rad/s)

b).

The emf alternates sinusoidally between +20mV and -20mV, so strictly speaking the average emf is zero. In AC work, as in other cases when the average is zero, it is more informative to use another kind of "average", such as the "root mean square" emf (square the emf, integrate over the period, divide by the period, take the square root). When the variation is sinusoidal, the rms value is 1/2=0.7071 times the peak emf.

So rms emf is 0.7071x20mV = 14.14mV.

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