JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Magnetic Properties of Matter

A short bar magnet placed with its axis at

3 0^{\circ}

with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is

α \times 1 0^{- 3} J

. The value of

α

is ______.

Answer: 64

Solution

The torque on the magnet is

τ = μ B sin θ

, so

0.016 = μ B sin 3 0^{\circ} = \frac{1}{2} μ B

, giving

μ B = 0.032

N·m. The most stable orientation is

θ = 0

and the most unstable is

θ = 18 0^{\circ}

, so the work done is

W = U (18 0^{\circ}) - U (0^{\circ}) = - μ B cos 18 0^{\circ} + μ B cos 0^{\circ} = 2 μ B = 0.064

J. Writing

0.064 = 64 \times 1 0^{- 3}

J gives

α = 64

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Integer answer