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  • A   long conductor carrying a current of <img alt="50 \mathrm{~A}" src="h

A  0.1 \mathrm{~m} long conductor carrying a current of 50 \mathrm{~A} is held perpendicular to a magnetic field of 1.25 \mathrm{mT}. The mechanical power required to move the conductor with a speed of 1 \mathrm{~ms}^{-1} is:

Option: 1

62.5 mW


Option: 2

625 mW


Option: 3

6.25 mW


Option: 4

12.5 mW


Answers (1)

best_answer

Here, \mathrm{ l=0.1 \mathrm{~m}, \mathrm{v}=1 \mathrm{~ms}^{-1} }
\mathrm{ \mathrm{I}=50 \mathrm{~A}, \mathrm{~B}=1.25 \mathrm{mT}=1.25 \times 10^{-3} \mathrm{~T} }
The induced emf is \mathrm{ \varepsilon=\mathrm{B} / \mathrm{v} }
The mechanical power is
\mathrm{ \mathrm{P}=\varepsilon \mathrm{I}=\mathrm{Bl} \mathrm{vI}=1.25 \times 10^{-3} \times 0.1 \times 1 \times 50=6.25 \times 10^{-3} \mathrm{~W}=6.25 \mathrm{~mW} }

Posted by

Deependra Verma

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