Question

A semi circular current carrying wire having radius R is placed in x-yplane with its centre at origin 'O'. There is non-uniform magnetic field $\overrightarrow{B}=\frac{B_{o}x}{2R}\hat{k}$(here $B_o$ is $+$ve constant) is existing in the region. The magnetic force acting on semi circular wire will be along.

• A. $-$x-axis
• B. $+$y-axis
• C. $-$y-axis
• D. $+$x-axis

Answer 1

The correct option is A $-$x-axis

Consider a small element of curve of length $Rd\theta$ with current I direction of $\overrightarrow{dl}$ can be written as

$\overrightarrow{dl}$ $Rd\theta$

$(sin\theta \hat{i}+cos\theta \hat{j})$

magnetic field acting on $dl$

$=\frac{B_0}{2R}\ast (-Rcos\theta )\hat{k}$

$\therefore$force acting on small part

$dF=I(\overrightarrow{dl}\ast \overrightarrow{B})$

$=IRd\theta (sin\theta \hat{i}+cos\theta \hat{j})\ast \left(\frac{B_{0}Rcos\theta }{2R}\hat{k}\right)$

$=\frac{I{B}_{0}R}{2}[sin\theta cos\theta \hat{j}-co{s}^2\theta \hat{i}]d\theta$

$\overrightarrow{F_{total}}={\int }_0^{\pi }dF=\frac{I{B}_{0}R}{2}{\int }_0^{\pi }[\frac{sin2\theta }{2}\hat{j}-\frac{(1+cos2\theta )}{2}\hat{i}]d\theta$

$=\frac{I{B}_{0}R}{2}[0-\frac{\pi }{2}\hat{i}-0]=\frac{\pi I{B}_{0}R}{4}(-\hat{i})$

Hence along -ve x axis