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3 years ago

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A 16.0 m long, thin, uniform metal rod slides north at a speed of 21.0 m/s. The length of the rod maintains an east-west orienta

tion while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 42.0 µT. What is the magnitude of the induced emf between the ends of the rod (in mV)?

1 answer:

Gekata [30.6K]3 years ago

4 0

Answer:

14.112 mV

Explanation:

L = 16 m, v = 21 m/s, B = 42 μ T = 42 x 10^-6 T

The formula for the induced emf is given by

e = B x v x L

e = 42 x 10^-6 x 21 x 16 = 14.112 x 10^-3 V = 14.112 mV

Thus, the induce emf is 14.112 mV.

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