Question

A bar magnet is initially parallel to the magnetic field in which it is placed. Work done to rotate it by ${60}^0$ is $2.6$ J. Work done in rotating it further by ${30}^0$ will be

• A. $1.3$J
• B. Zero
• C. $2.6$J
• D. $5.4$ J

Answer 1

The correct option is C $2.6$J

$\text{We Know,}$
$U=-mB\cos \theta$

$W=U_f-U_i$

U=potential energy,
m= magnetic dipole moment,
B= magnetic field,
$\theta$= angle between $m$ and $B$.
$\text{In first case}\theta =0^0-{60}^0$

$W=\frac{m.B}{2}=2.6$ Joules

$\text{In second case}\theta ={60}^0-{90}^0$

$W=\frac{m.B}{2}$

And we know that,$\frac{m.B}{2}=2.6$ Joules.

Hence work done = $2.6$ Joules.

Hence option $\text{C}$ is correct answer.