Search any question & find its solution
Question:
Answered & Verified by Expert
A bar magnet having a magnetic movement of $2 \times 10^4 \mathrm{JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6 \times 10^{-4}$ $T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is :
Options:
Solution:
2131 Upvotes
Verified Answer
The correct answer is:
$6 \mathrm{~J}$
$$
\begin{aligned}
\mathrm{W} & =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =2 \times 10^4 \times 6 \times 10^{-4}\left(\cos \theta-\cos 60^{\circ}\right) \\
& =12 \times \frac{1}{2}=6 \mathrm{~J}
\end{aligned}
$$
\begin{aligned}
\mathrm{W} & =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =2 \times 10^4 \times 6 \times 10^{-4}\left(\cos \theta-\cos 60^{\circ}\right) \\
& =12 \times \frac{1}{2}=6 \mathrm{~J}
\end{aligned}
$$
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.