A small bar magnet placed with its axis at

Question:

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

 

 

  1. $9.2 \times 10^{-3} \mathrm{~J}$

  2. $6.4 \times 10^{-2} \mathrm{~J}$

  3. $11.7 \times 10^{-3} \mathrm{~J}$

  4. $7.2 \times 10^{-2} \mathrm{~J}$


Correct Option: , 4

Solution:

Torque on a bar magnet : I = MB $\sin \theta$

Here, $\theta=30^{\circ}, \mathrm{I}=0.018 \mathrm{~N}-\mathrm{m}, \mathrm{B}=0.06 \mathrm{~T}$

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

$\Rightarrow 0.018=\mathrm{M} \times 0.06 \times \frac{1}{2}$

$\Rightarrow \mathrm{M}=0.6 \mathrm{~A}-\mathrm{m}^{2}$

Now $\mathrm{v}=-\mathrm{MB} \cos \theta$

Position of stable equilibrium $\left(\theta=0^{\circ}\right)$ :

$\mathrm{u}_{\mathrm{i}}=-\mathrm{MB}$

Position of unstable equilibrium $\left(\theta=180^{\circ}\right)$ :

$\mathrm{u}_{\mathrm{f}}=\mathrm{MB}$

$\Rightarrow$ work done $: \Delta \mathrm{U}$

$\Rightarrow \mathrm{W}=2 \mathrm{MB}$

$\Rightarrow \mathrm{W}=2 \times 0.6 \times 0.06$

$\Rightarrow \mathrm{W}=7.2 \times 10^{-2} \mathrm{~J}$

option (4) is correct

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