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

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 (A) 4.8× 10- 2J (B) 9.6× 10- 2J (C) 2.8× 10- 2J (D) 1.2× 10- 2J

Tardigrade Admin · Accepted Answer

For the unstable equilibrium, the angle between the magnetic moment and magnetic field is 180^o. (

∵

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

θ = 18 0^{o}

Potential energy of the magnet
U = - mB cos 180^o
= - 0.32 x 0.15 (- 1) = 4.8 x 10^-2 J
Thus, for the unstable equilibrium the potential energy is 4.8 x 10^-2 J.

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

$4.8 \times 1 0^{- 2} J$

$9.6 \times 1 0^{- 2} J$

$2.8 \times 1 0^{- 2} J$

$1.2 \times 1 0^{- 2} J$

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

4.8×10−2J

9.6×10−2J

2.8×10−2J

1.2×10−2J

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

$4.8 \times 1 0^{- 2} J$

$9.6 \times 1 0^{- 2} J$

$2.8 \times 1 0^{- 2} J$

$1.2 \times 1 0^{- 2} J$