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Magnetism and Matter

Section: Physics

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Question : 50 of 75

Marks: +1, -0

A small bar magnet placed with its axis at

30^{\circ}

with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

[4 Sep 2020 Shift 1]

$9.2 \times 10^{-3} \text{J}$
$6.4 \times 10^{-2} \text{J}$
$11.7 \times 10^{-3} \text{J}$
$7.2 \times 10^{-2} \text{J}$

Solution:

Torque on a bar magnet

: I = MB \sin \theta

Here,

\theta = 30^{\circ}, I = 0.018 \text{N} \cdot \text{m}, B = 0.06 \text{T}

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

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

\Rightarrow M = 0.6 \text{A} \cdot \text{m}^{2}

Now

v = -MB \cos \theta

Position of stable equilibrium

(\theta = 0^{\circ}):

u_i = -MB

Position of unstable equilibrium

(\theta = 180^{\circ}):

u_f = MB

\Rightarrow

work done

:\Delta U

\Rightarrow W = 2 MB

\Rightarrow W = 2 \times 0.6 \times 0.06

\Rightarrow W = 7.2 \times 10^{-2} \text{J}

option (4) is correct

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