Question

A short bar magnet, placed with its axis at 30° to a uniform external magnetic field of 0.16T, experience a torque of magnitude 0.032 N–m. Its potential energy, in the field, for its two positions of equilibriums, is

(1) 0.064 J for both stable and unstable equilibrium.

(2) 0.064 J for stable equilibrium and zero for unstable equilibruim

(3) 0.064 J for stable equilibrium and –0.064 J for unstable equilibrium

(4) –0.064 for stable equilibrium and +0.064 for unstable equilibrium

Answer 1

(4) –0.064 for stable equilibrium and +0.064 for unstable equilibrium

Potential energy of a bar magnet is given by, V = mBcosθ

We are given that

For stable equilibrium θ = 0 and for unstable equilibrium θ = π

Subsituting the values, the P.E. V – 0.064 J for stable equilibrium and + 0.064 J for unstable equilibrium.