Physics - Magnetism Question with Solution | TestHub

Let the radius of the circular coil be $r$ and the current be $I$.

The area of the coil is $A=\pi r^2$, which gives $r=\frac{A^{1/2}}{{\pi }^{1/2}}$.

The magnetic field at the centre of the circular coil is $B=\frac{{\mu }_{0}I}{2r}$.

From this, the current is $I=\frac{2rB}{{\mu }_0}$.

The magnetic moment of the coil is $M=IA$.

Substituting the value of $I$, we get $M=\left(\frac{2rB}{{\mu }_0}\right)A$.

Substituting $r=\frac{A^{1/2}}{{\pi }^{1/2}}$, we get $M=\frac{2BA}{{\mu }_0}\times \frac{A^{1/2}}{{\pi }^{1/2}}=\frac{2B{A}^{3/2}}{{\mu }_0{\pi }^{1/2}}$.

Answer: $\frac{2B{\mathrm{A}}^{3/2}}{{\mu }_0{\pi }^{1/2}}$