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  • A 0.1 m long conductor 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 <img alt="\mathrm{1 \; ms

A 0.1 m long conductor 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 \mathrm{1 \; 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

\mathrm{}Here, l=0.1 \mathrm{~m}, \mathrm{v}=\mathrm{ms}^{-1}

\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{\begin{aligned} & \mathrm{P}=e \mathrm{I}=\mathrm{BlvI}=1.25 \times 10^{-3} \times 0.1 \times 1 \times 50 \\ & =6.25 \times 10^{-3} \mathrm{~W}=6.25 \mathrm{~mW} \end{aligned}}

Posted by

Pankaj

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