A short bar magnet placed with its axis at 30° 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 × 10⁻³ J. The value of α i s _______.

Question

G1 Physics · Accepted Answer

Torque on a magnetic dipole is τ = μB sinθ. Given, τ = 0.016 N·m, θ = 30°. Therefore, 0.016 = μB(1/2). Hence, μB = 0.032 J. Potential energy of a magnetic dipole is U = -μB cosθ. Most stable position: θ = 0°, U(stable) = -μB. Most unstable position: θ = 180°, U(unstable) = +μB. Therefore the required external work is W = U(unstable)-U(stable) = μB-(-μB) = 2μB = 2×0.032 = 0.064 J = 64×10⁻³ J. Hence, α = 64.