Question

The torque required to keep a magnet of length $20\text{cm}$ at ${30}^{\circ }$ to a uniform magnetic field is $2\times {10}^{-5}\text{Nm}$. The magnetic force on each pole is-

• A. $2\times {10}^{-6}\text{N}$
• B. $2\times {10}^{-5}\text{N}$
• C. $2\times {10}^{-4}\text{N}$
• D. $2\times {10}^{-3}\text{N}$

Answer 1

The correct option is C $2\times {10}^{-4}\text{N}$

Given, $2l=20\text{cm};\tau =2\times {10}^{-5}\text{Nm}$

The torque required to keep the magnet at ${30}^{\circ }$ is,

$\tau =MB\sin \theta [\because M=m\times 2l]$

$\tau =m\times 2l\times B\sin {30}^{\circ }$

$2\times {10}^{-5}=mB\times 20\times {10}^{-2}\times \frac{1}{2}$

Force on each pole is, $F=mB$

$\Rightarrow mB=\frac{2\times {10}^{-5}}{10\times {10}^{-2}}$

$=2\times {10}^{-4}\text{N}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.