Question

The rigid conducting thin wire frame carries an electric current $I$ and this frame is inside a uniform magnetic field $\overrightarrow{B}$ as shown in fig. Then,

• A. the net magnetic force on the frame is zero but the torque is not zero
• B. the net magnetic force on the frame and the torque due to magnetic field are both zero
• C. the net magnetic force on the frame is not zero and the torque is also not zero
• D. none of these

Answer 1

The correct option is A the net magnetic force on the frame is zero but the torque is not zero

The forces on the sides $PQ,QR$ and $RP$ are outward.

Let the lengths of $PQ=RP=a$

$\therefore QR=\sqrt{2}a$ since $PQR$ is a right angle triangle.

$\overrightarrow{F_{PQ}}=-IaB\hat{i}$ where we have used $\overrightarrow{F}=\int I(\overrightarrow{dl}\times \overrightarrow{B})$

$\overrightarrow{F_{RP}}=-IaB\hat{j}$

$\overrightarrow{F_{QR}}=I\left(\sqrt{2}aB\right)\frac{(\hat{i}+\hat{j})}{\sqrt{2}}$

Net force on the frame ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_{PQ}+{\overrightarrow{F}}_{RP}+{\overrightarrow{F}}_{QR}=-IaB\hat{i}-IaB\hat{j}+I\left(\sqrt{2}aB\right)\frac{(\hat{i}+\hat{j})}{\sqrt{2}}=0$
Logically one can see this without doing the calculation
$\overrightarrow{QR}$ resolves x-component and y-component which have current in the opposite direction to RP and PQ respectively.
Hence B produced by them will mutually cancel pairwise.

Net torque $\tau =\overrightarrow{\mu }\times \overrightarrow{B}$

$\left|\tau \right|=\left|\overrightarrow{\mu }\right|\left|\overrightarrow{B}\right|\sin \theta =IAB\sin {90}^0=IAB$ where $A$ is the area of the triangle.

Thus, $\left|\tau \right|$ is non-zero