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 $1.5Wb{m}^{-2}$ through ${180}^o$ about an axis perpendicular to the magnetic field?

Answer 1

The potential energy of circular loop in the magnetic field is $-MBcos\theta$, where $\theta$ is angle between normal to plane of coil and B. Initially given $\theta =0^o$
$\therefore$ Initial potential energy, $U_i=NiABcos{0}^o=-NiAB$
When coil is turned through ${180}^o;\theta ={180}^o$, therefore, final potential energy
$U_f=-NiABcos{180}^o=NiAB$
$\therefore$ Required work, W $=$ gain in potential energy
$=U_f-U_i=NiAB-(-NiAB)=2NiAB$

Here, $N=100,i=0.1A,B=1.5wb/m^2$ and radius $r=0.05m$
$\therefore A=\pi r^2=3.14\times (0.05{)}^2=7.85\times {10}^{-3}{m}^2$
$\therefore Work,W=2\times 100\times 0.1\times 7.85\times {10}^{-3}\times 1.5=0.2355J$