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A 0.76-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 25 m/s, and the emf induced across its length is 7.10 10-4 V. Assume the horizontal component of the earth's magnetic field at the location of the bar points directly north.

Respuesta :

Answer:

3.74 x 10^-5 T

Explanation:

length, l = 0.76 m

velocity, v = 25 m/s

induced emf, e = 7.10 x 10^-4 V

Let B be the magnetic field.

According to the formula

induced emf, e = B x v x l

7.10 x 10^-4 = B x 25 x 0.76

B = 3.74 x 10^-5 T

Thus, the magnetic field 3.74 x 10^-5 T.

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