A magnet of magnetic moment $50\hat iAm^2$ is placed along the X-axis in a magnetic field $\vec B=(0.5\hat i+3.0\hat j) T$. The torque acting on it is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $150\hat k\, Nm$
The torque ($τ$) acting on a magnetic dipole is,
$τ=\vec m×\vec B$
$=(50\hat i)×(0.5\hat i+3.0\hat j)$
and,
$\hat i×\hat i=0$
$\hat i×\hat j=k$
$τ=50×[\hat i(0.5\hat i+3.0\hat j)]$
$=50×3\hat k$
$=150\hat kNm$