Search any question & find its solution
Question:
Answered & Verified by Expert
A magnet of magnetic moment $2 \mathrm{~J} \mathrm{~T}^{-1}$ is aligned in the direction of magnetic field of $0.1 \mathrm{~T}$. What is the net work done to bring the magnet normal to the magnetic field?
Options:
Solution:
2264 Upvotes
Verified Answer
The correct answer is:
$0.2 \mathrm{~J}$
Given that, magnetic moment, $M=2 \mathrm{JT}^{-1}$
Magnetic field, $B=0.1 \mathrm{~T}$
Since, magnetic moment is aligned in the direction of magnetic field, i.e. initial angle,
$$
\theta_1=0^{\circ}
$$
When magnet is normal, then final angle, $\theta_2=90^{\circ}$ Using relation for work done,
$$
W=-M B\left(\cos \theta_2-\cos \theta_1\right)
$$
Substituting the values, we get
$$
\begin{aligned}
W & =-2 \times 0.1\left(\cos 90^{\circ}-\cos 0^{\circ}\right) \\
& =-2 \times 0.1(0-1)=0.2 \mathrm{~J}
\end{aligned}
$$
Hence, $0.2 \mathrm{~J}$ work will be done to rotate a magnet from $0^{\circ}$ to $90^{\circ}$ angle with in magnetic field.
Magnetic field, $B=0.1 \mathrm{~T}$
Since, magnetic moment is aligned in the direction of magnetic field, i.e. initial angle,
$$
\theta_1=0^{\circ}
$$
When magnet is normal, then final angle, $\theta_2=90^{\circ}$ Using relation for work done,
$$
W=-M B\left(\cos \theta_2-\cos \theta_1\right)
$$
Substituting the values, we get
$$
\begin{aligned}
W & =-2 \times 0.1\left(\cos 90^{\circ}-\cos 0^{\circ}\right) \\
& =-2 \times 0.1(0-1)=0.2 \mathrm{~J}
\end{aligned}
$$
Hence, $0.2 \mathrm{~J}$ work will be done to rotate a magnet from $0^{\circ}$ to $90^{\circ}$ angle with in magnetic field.
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.