Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1506 Upvotes
Verified Answer
The correct answer is:
\(2 \pi \mathrm{J}\)
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).
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).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.