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 ${180}^o$ about its axis perpendicular to the magnetic field? The plane of the coil is initially perpendicular to the magnetic field :

• A. $0.456J$
• B. $2.65J$
• C. $0.2355J$
• D. $3.95J$

Answer 1

The correct option is C $0.2355J$

The potential energy of circular loop in the magnetic field $=-MB\cos \theta$, where $\theta$ is the angle between normal to plane of coil and magnetic field $B$.

Initial potential energy,

$U_1=-NiAB\cos 0^o=-NiAB$

When coil is turned through ${180}^o$, therefore final potential energy,

$U_f=-NiAB\cos {180}^o=NiAB$

$\therefore$ Required work, $W=$ gain in potential energy

$=U_i-U_i=NiAB-(-NiAB)=2NiAB$

Here, $N=100$, $i=0.1A$, $B=1.5Wb/m^2$ and radius $r=0.05m$

$\therefore A=\pi r^2=3.14\times {\left(0.05\right)}^2=7.85\times {10}^{-3}{m}^2$

$\therefore$ Work, $W=2\times 100\times 0.1\times 7.85\times {10}^{-3}\times 1.5$

$\Rightarrow W=0.2355J$