Question

The plane of a rectangular loop of wire with sides $0.05m$ and $0.08m$ is parallel to a uniform magnetic field of induction $1.5\times {10}^{-2}T$ . 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 lines of field, then the torque acting on it is

• A. $6000N-m$
• B. $zero$
• C. $1.2\times {10}^{-2}N-m$
• D. $6\times {10}^{-4}N-m$

Answer 1

Answer: (D)

$6\times {10}^{-4}N-m$

Torque on the loop $\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$ where $\overrightarrow{\mu }=i\overrightarrow{A}$ is the magnetic moment of the loop and $\overrightarrow{B}$ is the magnetic field.
$\therefore \tau =i\overrightarrow{A}\times \overrightarrow{B}$
Since $\overrightarrow{B}$ is in the plane of the loop, $\overrightarrow{A}\perp \overrightarrow{B}$
$\therefore \overrightarrow{\tau }=10\times (0.05\times 0.08)\times (1.5\times {10}^{-2})=6\times {10}^{-4}N-m$