Question

A circular loop of radius $R$ carries a current $I$ . Another circular loop of radius $r(<<R)$ carries a current $i$ and is placed at the centre of the larger loop . The planes of the two circles are at right angle to each other. Find the torque acting on the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum mangnitude of this force?

Answer 1

The torque on the smaller loop is given as,

$T=i\left(\overrightarrow{A}\times \overrightarrow{B}\right)$

$=i\times \pi r^2\times \frac{{\mu }_{0}I}{2r}\sin {90}^{\circ }$

$=\frac{iI{\mu }_0\pi r}{2}$

Thus, acting torque on the smaller loop is $\frac{iI{\mu }_0\pi r}{2}$.