Question

A small current carrying loop having current $i$ is placed in the plane of the paper as shown in figure. Another semicircular loop having current $i_0$ is placed concentrically in the same plane as that of the small loop, the radius of semicircular loop being $R(R>>a)$. Find the force applied (in newton) by the smaller ring on the bigger ring.
(Given $R=1m,i=i_0=\frac{40}{\sqrt{{\mu }_0}}A,a=0.1m$)

Answer 1

Magnetic field due to smaller ring at a distance $R$ (equatorial point) from the centre:

$B=\frac{{\mu }_0}{4\pi }\frac{i\pi a^3}{R^3}=\frac{{\mu }_{0}i\pi a^2}{{4R}^3}$ where $M=i\pi a^2$
$B=\frac{{\mu }_0}{4\pi }\frac{i\pi a^3}{R^3}=\frac{{\mu }_{0}i\pi a^2}{{4R}^3}$
$dF=i_{0}Rd\theta \frac{{\mu }_{0}i{a}^2}{{4R}^3}$
${dF}_x=dF\sin \theta =\frac{{\mu }_{0}i{i}_0{a}^2\sin \theta d\theta }{{4R}^2}$
$F_x=\frac{{\mu }_{0}i{i}_0{a}^2}{{4R}^2}{\int }_0^{\pi }\sin \theta d\theta$
$\Rightarrow F_x=\frac{{\mu }_{0}i{i}_0{a}^2}{{4R}^2}\times 2=\frac{{\mu }_{0}i{i}_0{a}^2}{{2R}^2};F_y=0$
$\therefore F_{net}=F_x=\frac{{\mu }_{0}i{i}_0{a}^2}{{2R}^2}=8N$