Question

A bar magnet of magnetic moment 1.5 J/T is along the direction of the uniform magnetic field of 0.22T. The work done in turning the magnet opposite to the field direction and the torque required to keep in that position are

• A. $0.33J$ and $0.33N-m$
• B. $0.66J$ and $0.66N-m$
• C. $0.33J$ and $0N-m$
• D. $0.66J$ and $0N-m$

Answer 1

Answer: (D)

$0.66J$ and $0N-m$

Given:

Magnetic moment =1.5 J/T

B= 0.22T

Solution:

Work done, W= - $mB(Cos{\theta }_2-Cos{\theta }_1)$

Here, ${\theta }_1=0°and{\theta }_2=180°$

Therefore, W=-1.5×0.22×(Cos180° -Cos0°)

$\Rightarrow W=-1.5\times 0.22\times (-1-1)$

$\Rightarrow W=0.66J$

Torque,T= mBSin$\theta$

$\Rightarrow T=1.5\times 0.22\times Sin180°$

Therefore, T = 0 Nm since Sin180° =0

Hence, correct option is D.