A circular coil with cross-sectional area 0.2 cm^2 carries a current of 4 A. It is kept in a uniform magnetic field of magnitude 0.5 T normal to the…

(a) Net force = zero By symmetry, force on each element of the coil is equal and opposite to the force on the diametrically opposite element of the coil. Hence, the net force is zero. Alternatively : \[\begin{aligned} & d \vec{f}_{1}=I \overrightarrow{d l} \times \vec{B} \\ & d \vec{f}_{2}=I \overrightarrow{d l} \times \vec{B} \\ & \vec{F}_{1}=-\vec{F}_{2} \end{aligned} \] On coil magnetic force are equal and opposite, so they will cancel out each other.

(b) Torque on the coil. \[\begin{aligned} & \vec{\tau}=\vec{m} \times \vec{B} \\ & \tau=m B \sin \theta \quad[\vec{m} \| \vec{B}] \\ & \tau=m B \sin 0^{\circ} \quad \theta=0^{\circ} \\ & \tau=0 \end{aligned} \]

(c) \[\begin{aligned} f_{\text {avg }} & =e\left(\overrightarrow{\mathrm{v}}_{d} \times \vec{B}\right) \\ & =e \mathrm{v}_{d} B \sin 90^{\circ} \\ & =e \mathrm{v}_{d} B \end{aligned} \] \[\therefore I=n e A \mathrm{v}_{d} \]