A circular coil of radius \(10 \mathrm{~cm}\) with 100 turns carrying a current of \(0.5 \mathrm{~A}\) lies…
A circular coil of radius \(10 \mathrm{~cm}\) with 100 turns carrying a current of \(0.5 \mathrm{~A}\) lies in a magnetic field of \(2 \mathrm{~T}\) such that the normal drawn to the plane of the coil makes an angles \(\theta\) with the direction of the field. Work done in rotating the coil to change the angle \(\theta\) from \(0^{\circ}\) to \(180^{\circ}\) is
\(\pi \mathrm{J}\)
\(2 \pi \mathrm{J}\)
\(4 \pi \mathrm{J}\)
\(8 \pi \mathrm{J}\)
Solution
Given, radius of coil, \(R=10 \times 10^{-2} \mathrm{~m}\), number of turns in coil, \(N=100\) turns, current through coil, \(I=0.5 \mathrm{~A}\), magnetic field, \(B=2 \mathrm{~T}\) and change in angle of rotating coil, \(\theta=0^{\circ}\) to \(180^{\circ}\).
Work done in turning a loop from angle \(\theta_1\) to \(\theta_2\).
\(\begin{aligned}
& W=M B\left(\cos \theta_1-\cos \theta_2\right) \\
\Rightarrow \quad & W=N I A B\left[\cos 0^{\circ}-\cos 180^{\circ}\right] \\
\Rightarrow \quad & W=100 \times 0.5 \times \pi \times 100 \times 10^{-4} \times 2[1-(-1)]
\end{aligned}\)
Hence, \(W=2 \pi \mathrm{J}\)
\(\therefore\) The correct option is (b).