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Question

Physics Question on Magnetism and matter

A current flows in a conducting wire of length L. If we bend it in a circular form, its magnetic dipole moment would be

A

IL24π\frac{IL^2}{4\pi}

B

IL4π\frac{IL}{4\pi}

C

I2L4π\frac{I^2L}{4\pi}

D

I2L24π\frac{I^2L^2}{4\pi}

Answer

IL24π\frac{IL^2}{4\pi}

Explanation

Solution

Let a wire of length L is bend in a circular form of radius r Then 2πr=L 2\pi r=L \Rightarrow r=\frac{L}{2\pi} \hspace5mm ...(i) The magnetic dipole moment of a circular ring M=IA  M=IA\,\,[ A is area of the ring] or M=I\pi r^2\hspace5mm ...(ii) On putting the value of r from E (i) in E (ii), we have M=Iπ(L2π)2⇒M=Iπ×L24π2 M=I\pi \Bigg(\frac{L}{2\pi}\Bigg)^2 \Rightarrow M=I\pi \times\frac{L^2}{4\pi^2} ⇒   M=IL24π\Rightarrow \, \, \, M=\frac{IL^2}{4\pi}