What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10-5 T magnetic field, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 12.0 ms?

Question

Rogelio Allen

Physics

4 July, 19:55

What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10-5 T magnetic field, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 12.0 ms?

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Answers (1)

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Marvin · Answer 1

5.77 Volt

Explanation:

N = 1000, Diameter = 42 cm = 0.42 m, t = 12 ms = 12 x 10^-3 s

Change in magnetic field, B = 5 x 10^-5 T

The peak value of emf is given by

e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt