Q. A magnet of magnetic moment $50\widehat{i}\ A$-$m^{2}$ is placed along the $x$-axis in a magnetic field $\overrightarrow{B} = (0.5\ \widehat{i} + 3.0\widehat{j})T$. The torque acting on the magnet is
A) $175\ \widehat{k}\ N$-$m$
B) $150\ \widehat{k}\ N$-$m$
C) $75\ \widehat{k}\ N$-$m$
D) $25\sqrt{37}\ \widehat{k}\ N$-$m$
Explanation
(b) $\overrightarrow{\tau} = \overrightarrow{M} \times \overrightarrow{B} \Rightarrow \overrightarrow{\tau} = 50\widehat{i} \times (0.5\widehat{i} + 3\widehat{j})$
$= 150\left( \widehat{i} \times \widehat{j} \right) = 150\widehat{k}N \times m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.