In a uniform magnetic field of strength 0.15 T, a short bar magnet of magnetic moment m=0.32 J T -1 is placed. The potential energy of the magnet in its unstable equilibrium position is n × 10-23 then n will be:

Question

In a uniform magnetic field of strength 0.15 T, a short bar magnet of magnetic moment m=0.32 J T -1 is placed. The potential energy of the magnet in its unstable equilibrium position is n × 10-23 then n will be: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For the unstable equilibrium, the angle between the magnetic moment and magnetic field is

18 0^{\circ}

.
(

∵

In this position it will be in a direction perpendicular to magnetic field thus maximum torque will act on it.)

θ = 18 0^{\circ}

Potential energy of the magnet

U = - m B cos 18 0^{\circ}

= 0.32 \times 0.15 (- 1) = 4.8 \times 1 0^{- 2} J

Thus, for the unstable equilibrium the potential energy is

4.8 \times 1 0^{- 2} J

Q. In a uniform magnetic field of strength 0.15T, a short bar magnet of magnetic moment m=0.32JT−1 is placed. The potential energy of the magnet in its unstable equilibrium position is n×10−23 then n will be:

Q. In a uniform magnetic field of strength $0.15 T$ , a short bar magnet of magnetic moment $m = 0.32 J T^{- 1}$ is placed. The potential energy of the magnet in its unstable equilibrium position is $n \times 1 0^{- 23}$ then $n$ will be: