An imaginary north pole of 10 , Am is rotatin | vedclass

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Question

(B) The magnetic potential difference between two points in a magnetic field is given by the work done in moving a unit north pole. For a closed loop

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$\pi 	imes 10^{-5} \, J$

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(B) The magnetic potential difference between two points in a magnetic field is given by the work done in moving a unit north pole. For a closed loop around a current-carrying wire, the work done $W$ is given by $W = m \oint \vec{B} \cdot d\vec{l}$, where $m$ is the pole strength.From Ampere's Circuital Law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$.Therefore, the work done in one complete revolution is $W_{rev} = m \mu_0 I$.Given: $m = 10 \, Am$, $I = 5 \, A$, and $\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$.$W_{rev} = 10 	imes (4\pi 	imes 10^{-7}) 	imes 5 = 200\pi 	imes 10^{-7} = 2\pi 	imes 10^{-5} \, J$.The frequency of revolution is $30 \, 	ext{rpm} = 0.5 \, 	ext{revolutions/sec}$.Work done in one second = $(	ext{Work per revolution}) 	imes (	ext{Number of revolutions per second}) = (2\pi 	imes 10^{-5} \, J) 	imes 0.5 = \pi 	imes 10^{-5} \, J$.

An imaginary north pole of $10 \, Am$ is rotating around an infinitely long current-carrying wire at $30 \, \text{revolutions/min}$ on a circular path. If the current in the wire is $5 \, A$, then calculate the work done in one second.

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