A 0.1 m long conductor carrying a current of 50 A is held perpendicular to a magnetic field of 1.25

Question

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 $1 \mathrm{~ms}^{-1}$ is\

Professor PC Thomas Classes · Answer

Here, $\ell=0.1 \mathrm{~m}, \mathrm{v}=1 \mathrm{~ms}^{-1}$\ $\mathrm{I}=50 \mathrm{~A}, \mathrm{~B}=1.25 \mathrm{mT}=1.25 	imes 10^{-3} \mathrm{~T}$\ The induced emf is $\varepsilon=\mathrm{B} \ell \mathrm{v}$\ The mechanical power is\ $\mathrm{P}=\varepsilon \mathrm{I}=\mathrm{B} \ell \mathrm{VI}=1.25 	imes 10^{-3} 	imes 0.1 	imes 1 	imes 50$\ $=6.25 	imes 10^{-3} \mathrm{w}=6.25 \mathrm{~mW}$\