A bar magnet of length $10 \mathrm{~cm}$ and having the pole strength equal to $10^{-3}…

A bar magnet of length $10 \mathrm{~cm}$ and having the pole strength equal to $10^{-3} \mathrm{~A}-\mathrm{m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \mathrm{~T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is
  1. $2 \pi \times 10^{-7} \mathrm{Nm}$
  2. $2 \pi \times 10^{-5} \mathrm{Nm}$
  3. $0.5 \mathrm{Nm}$
  4. $0.5 \times 10^2 \mathrm{Nm}$

Solution

Given, length of bar magnet, $ l=10 \mathrm{~cm}=10^{-1} \mathrm{~m} $ Pole strength, $m=10^{-3} \mathrm{~A}-\mathrm{m}$ $ \begin{aligned} & B=4 \pi \times 10^{-3} \mathrm{~T} \\ & \theta=30^{\circ} \end{aligned} $ $\therefore$ Magnetic dipole moment, $ M=m l=10^{-3} \times 10^{-1}=10^{-4} \mathrm{~A}-\mathrm{m}^2 $ $\therefore$ Torque, $\tau=M B \sin \theta$ $ \begin{aligned} & =10^{-4} \times 4 \pi \times 10^{-3} \times \sin 30^{\circ} \\ & =4 \pi \times 10^{-7} \times \frac{1}{2}=2 \pi \times 10^{-7} \mathrm{~N}-\mathrm{m} \end{aligned} $

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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