JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Magnetic Properties of Matter

A small bar magnet placed with its axis at

3 0^{\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:

(A) $11.7 \times 1 0^{- 3} J$
(B) $9.2 \times 1 0^{- 3} J$
(C) $7.2 \times 1 0^{- 2} J$ ✓ Correct
(D) $6.4 \times 1 0^{- 2} J$

Answer: (C)

Solution

The torque fixes the magnetic moment:

τ = μ B sin θ

, so

0.018 = μ \times 0.06 \times sin 3 0^{\circ} = μ \times 0.06 \times 0.5

, giving

μ = 0.6

\cdot

^{2}

. Carrying the magnet from stable equilibrium (

θ = 0^{\circ}

) to unstable equilibrium (

θ = 18 0^{\circ}

) needs work equal to the change in orientation energy

U = - μ B cos θ

W = (- μ B cos 18 0^{\circ}) - (- μ B cos 0^{\circ}) = 2 μ B = 2 \times 0.6 \times 0.06 = 0.072

J, that is

7.2 \times 1 0^{- 2}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Magnetic Properties of Matter

Solution

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