Question

A small magnet placed with is axis at ${30}^{\circ }$ with an external field of $0.06\text{T}$ experience a torques of $0.018\text{Nm}.$ The minimum work to rotate it from its stable to unstable equilibrium position is

• A. $6.4\times {10}^{-2}\text{J}$
• B. $9.2\times {10}^{-3}\text{J}$
• C. $7.2\times {10}^{-2}\text{J}$
• D. $11.7\times {10}^{-3}\text{J}$

Answer 1

The correct option is C $7.2\times {10}^{-2}\text{J}$

Given, $\theta ={30}^{\circ };B=0.06\text{T}\tau =0.018\text{Nm}$

$\because \tau =MB\sin \theta$

$0.018=M\times 0.06\times \sin {30}^{\circ }$

$\Rightarrow M=0.6{\text{Am}}^2$

$W={\int }_{{\theta }_1}^{{\theta }_2}\tau d\theta ={\int }_{{\theta }_1}^{{\theta }_2}MB\sin \theta d\theta [\because {\theta }_1=0^{\circ };{\theta }_2={180}^{\circ }]$

$\Rightarrow W=-MB\cos \theta {|}_{{\theta }_1}^{{\theta }_2}=MB(\cos {\theta }_1-\cos {\theta }_2)$

$\Rightarrow W=2MB=2\times 0.6\times 0.06=7.2\times {10}^{-2}\text{J}$

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