A magnet of magnetic moment $2 \mathrm{~J} \mathrm{~T}^{-1}$ is aligned in the direction of magnetic field…
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?
$0.1 \mathrm{~J}$
$0.2 \mathrm{~J}$
$1.0 \mathrm{~J}$
$2.0 \mathrm{~J}$
Solution
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