Question
A circular coil of 100 turns and of diameter $20 \mathrm{~cm}$ carries a current of $1 \mathrm{~A}$. It is to be turned to an angle $\theta$ in a magnetic field of $5 \mathrm{~T}$ from a position $\theta=0^{\circ}$ to $\theta=180^{\circ}$. The magnetic moment of the coil is(a) $\pi \mathrm{A}-\mathrm{m}^{2}$(b) $\frac{\pi}{2} \mathrm{~A}-\mathrm{m}^{2}$(c) $2 \pi \mathrm{A}-\mathrm{m}^{2}$(d) $\frac{3 \pi}{8} \mathrm{~A}-\mathrm{m}^{2}$
Step 1
Step 1: The magnetic moment (M) of a coil is given by the formula: \[ M = NIA \] where N is the number of turns in the coil, I is the current flowing through the coil, and A is the area of the coil. Show more…
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