JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

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 ______.

Your answer

Solution

Convert: $B = 800$ G $= 0.08$ T.

Find magnetic moment from torque: $τ = μ B sin θ$ .

$0.016 = μ \times 0.08 \times sin 3 0^{\circ} = μ \times 0.04 \Rightarrow μ = 0.4$ J/T.

Most stable to most unstable means rotating from $θ = 0$ (aligned) to $θ = 18 0^{\circ}$ (anti-aligned). Using $U = - μ B cos θ$ :

$W = U_{f} - U_{i} = (+ μ B) - (- μ B) = 2 μ B = 2 \times 0.4 \times 0.08 = 0.064 J = 64 \times 1 0^{- 3}$ J.

$∴ α = 64$ .

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