Question

A circular coil of $100$ turns has an effective radius of $0.05m$ and carries a current of $0.1A$. How much work is required to turn it in an external magnetic field of $1.5Wb/m^2$ 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 $=-MB\cos \theta$,
where $\theta$ is the angle between nomal to plane of coil and magnetic field $B$.
Initial potential energy,
$U_1=-NiAB\cos 0^{\circ }=-NiAB$
When coil is tumed through $180^{\circ }$,
therefore final potential energy,
$U_f=-NiAB\cos 180^{\circ }=NiAB$
$\therefore$ Required work, $W=$ gain in potential energy
$=U_f-U_1=NiAB-(-NiAB)=2NiAB$
Here, $N=100,i=0.1A,B=1.5Wb/m^2$ and radius $r=0.05m$
$\therefore A=\pi r^2=3.14\times (0.05{)}^2=7.85\times 10^{-3}{m}^2$
$\therefore$ Work, $W=2\times 100\times 0.1\times 7.85\times 10^{-3}\times 1.5$
$\Rightarrow W=0.2355J$