Question

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

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

Answer 1

Answer: (D)

Torque on a bar magnet : $I=MB\sin \theta$
Here, $\theta =30^{\circ },I=0.018N-m,B=0.06T$
$\Rightarrow 0.018=M\times 0.06\times \sin 30^{\circ }$
$\Rightarrow 0.018=M\times 0.06\times \frac{1}{2}$
$\Rightarrow M=0.6A-m^2$
Now $v=-MB\cos \theta$
Position of stable equilibrium $\left(\theta =0^{\circ }\right)$ $u_i=-MB$
Position of unstable equilibrium $\left(\theta =180^{\circ }\right)$ :
$u_f=MB$
$\Rightarrow$ work done $:\Delta U$
$\Rightarrow W=2MB$
$\Rightarrow W=2\times 0.6\times 0.06$
$\Rightarrow W=7.2\times 10^{-2}J$
option (4) is correct