A bar magnet of magnetic moment 3 J/T lies aligned with the directions of a uniform magnetic field of 0.2 T. The amount of work, required so as to align its magnetic moment opposite to the field direction is:

1.2 J

The correct answer is Option (1) → 1.2 J

The work required rotate a magnetic moment $\vec M$ in a magnetic field $(\vec B)$ is -

$W=-MB(\cos θ_2-\cos θ_1)$

$M=3J/T$

$B=0.2T$

$θ_1=0°,θ_2=180°$

$W=-(3)(0.2)(\cos 180-\cos 0)$

$=1.2J$