A circular loop of radius R carries a current I . A second circular...

The magnetic field produced by the larger loop at its centre is given by $B = \dfrac{\mu_0 I}{2R}$.

The magnetic dipole moment of the smaller loop is $m = iA = i(\pi r^2)$.

Since the planes of the two loops are at right angles, the angle between the magnetic field vector $\vec{B}$ and the magnetic dipole moment vector $\vec{m}$ is 90^{}^{{}^{\circ}}.

The torque acting on the smaller loop is \tau = mB \sin 90^{}^{{}^{\circ}}.

Substituting the values of $m$ and $B$, we get $\tau = (i\pi r^2) \left( \dfrac{\mu_0 I}{2R} \right) = \dfrac{\mu_0 \pi i I r^2}{2R}$.