Physics - Magnetism Question with Solution | TestHub

Length of the wire is $L$.

Let the side of the square coil be $a$. Since the wire is turned into a square coil of a single turn, the perimeter of the square is equal to the length of the wire.

$4a=L\Rightarrow a=\frac{L}{4}$

Area of the square coil is $A=a^2={\left(\frac{L}{4}\right)}^2=\frac{L^2}{16}$

The magnetic dipole moment of the coil is given by $M=IA=\frac{I{L}^2}{16}$

The torque acting on the coil in a magnetic field $B$ is $\tau =MB\sin \theta$

For maximum torque, $\theta =90^{\circ }$, which gives $\sin \theta =1$.

${\tau }_{\text{max}}=MB=\frac{IB{L}^2}{16}$

Answer: $\frac{{\mathrm{IBL}}^2}{16}$