A short bar magnet of magnetic moment $m = 0.32 J T^{- 1}$ is placed in a uniform magnetic field of $0.15 T$ . If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Question

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Answer 1

(a) In stable equilibrium, the bar magnet is aligned along the magnetic field,i.e., $θ = 0^{\circ}$ .
Potential energy= $- m B c o s θ = - 4.8 \times 1 0^{- 2} J$
(b) In unstable equilibrium, the bar magnet is aligned reversed to magnetic field, i.e., $θ = 18 0^{\circ}$ .
Potential energy= $- m B c o s θ = 4.8 \times 1 0^{2} J$

Answer 2

(a) In stable equilibrium, the bar magnet is aligned along the magnetic field,i.e.,

θ = 0^{\circ}

.

Potential energy=

- m B c o s θ = - 4.8 \times 1 0^{- 2} J

(b) In unstable equilibrium, the bar magnet is aligned reversed to magnetic field, i.e.,

θ = 18 0^{\circ}

.

Potential energy=

- m B c o s θ = 4.8 \times 1 0^{2} J