Question

A circular coil of $100$ turns has an effective radius of $0.05\, m$ and carries a current of $0.1 \,A$. How much work is required to turn it in an external magnetic field of $1.5\, Wb / m ^{2}$ through $180^{\circ}$ about its axis perpendicular to the magnetic field? The plane of the coil is initially perpendicular to the magnetic field.

• A. 0,456 J
• B. 2.65 J
• C. 0.2355 J
• D. 3.95 J

Answer 1

Answer: (C)

The potential energy of circular loop in the magnetic field $=-M B \cos \theta$,
where $\theta$ is the angle between nomal to plane of coil and magnetic field $B$.
Initial potential energy,
$U_{1}=-N i A B \cos 0^{\circ}=-N i A B$
When coil is tumed through $180^{\circ}$,
therefore final potential energy,
$U_{f}=-N i A B \cos 180^{\circ}=N i A B$
$\therefore $ Required work, $W=$ gain in potential energy
$=U_{f}-U_{1}=N i A B-(-N i A B)=2 N i A B$
Here, $N=100, i=0.1 A , B=1.5 Wb / m ^{2}$ and radius $r=0.05 m$
$\therefore A=\pi r^{2}=3.14 \times(0.05)^{2}=7.85 \times 10^{-3} m ^{2}$
$\therefore $ Work, $ W=2 \times 100 \times 0.1 \times 7.85 \times 10^{-3} \times 1.5$
$\Rightarrow W=0.2355 \,J$