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A 0.1 m long coductor carrying a current...

A `0.1` m long coductor carrying a current of 50 A is held perpendicular to a magnetic field of `1.25` mT. The mechanical power required to move the conductor with a speed of `1 ms^(-1)` is

A

(a) `0.25 mW`

B

(b) `6.25mW`

C

( c) `0.625W`

D

(d) `1W`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `P = Fv = Bilv = 1.25 xx 10^(-3) xx 50 xx 0.1 xx 1 W`
`= 6,25 xx 10^(-3) W = 6.25m W`
An iterating alternating
`P = EI = (Blv) I`
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