Torque needed to deflect a magnet through $ 45^o$ in the earths horizontal magnetic field of induction $0.5\times 10^{-4}T$ is
(Given the magnetic moment of magnet is$ 4A-m^2$)
Answer & explanation
Correct answer: option 2
$ \tau = \vec{M} \times \vec{B} = MBsin45^o = 4 \times 0.5 \times 10^{-4} \times \frac{1}{\sqrt 2} = 1.41\times 10^{-4}$