Question

The plane of a rectangular loop of wire with sides $0.05m$ and $0.08m$ is parallel to a uniform magnetic field of $1.5\times {10}^{-2}T$. A current of $10.0A$ flows through the loop. If the side of length $0.08m$ is normal and the side of length $0.05m$ is parallel to the magnetic field, then the torque acting on the loop is

• A. $6000Nm$
• B. Zero
• C. $1.2\times {10}^{-2}Nm$
• D. $6\times {10}^{-4}Nm$

Answer 1

The correct option is D $6\times {10}^{-4}Nm$


Torque on a current carrying loop,
$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$
Magnitude of torque $\tau =MB\sin \theta$
Magnetic moment of a current carrying loop,
$M=NIA$ where $N$ is the number of turns of the loop, $I$ is the current flowing in the loop and $A$ is the area.
$M=1\times 10\times 8\times {10}^{-2}\times 5\times {10}^{-2}$ $=4\times {10}^{-2}A{m}^2$

$\therefore \tau =MB\sin \theta$
Since the area vector makes an angle of ${90}^{\circ }$ with the magnetic field, $\theta ={90}^{\circ }$
$\Rightarrow \tau =M\times B=4\times {10}^{-2}\times 1.5\times {10}^{-2}=6\times {10}^{-4}Nm$