Question

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

• A. $0.0235\text{J}$
• B. $0.0470\text{J}$
• C. $0.2355\text{J}$
• D. $0.4700\text{J}$

Answer 1

The correct option is C $0.2355\text{J}$

The potential energy of a circular loop in the magnetic field is,

$U=-\mu B\cos \theta$

Initially given, $\theta =0^{\circ }$

$\Rightarrow U_i=-NiAB\cos 0^{\circ }=-NiAB$

When coil is turned through ${180}^{\circ }$

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

Required external work=gain in potential energy

$\Rightarrow W=U_f-U_i=NiAB-(-NiAB)=2NiAB$

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

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

$\therefore W=2\times 100\times 0.1\times 7.85\times {10}^{-3}\times 1.5=0.2355\text{J}$

Hence, option (c) is the correct answer.