Question 195 of 276

A uniform magnetic field of $2\times {10}^{-3}T$ acts along positive Y-direction. A rectangular loop of sides $20cm$ and $10cm$ with current of $5A$ is in Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :

A$2\times {10}^{-4}Nm$ along positive Z –directionB$2\times {10}^{-4}Nm$ along negative Z-directionC$2\times {10}^{-4}Nm$ along positive X-directionD$2\times {10}^{-4}Nm$ along positive Y-direction

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Answer: (B) $2\times {10}^{-4}Nm$ along negative Z-direction

Magnetic moment is given by, $\vec{M}=i\vec{A}$
$=5\times \left(0.2\right)\times \left(0.1\right)\left(-\hat{i}\right)$
$=0.1\left(-\hat{i}\right)$
$\vec{\tau }=\vec{M}\times \vec{B}=0.1\left(-\hat{i}\right)\times \left(2\times {10}^{-3}\right)\left(\hat{j}\right)$
$=2\times {10}^{-4}\left(-\hat{k}\right)Nm$