Question

When the magnetic dipole is rotated through angle ${\displaystyle \theta }$, work done in rotating dipole is given by -mB cos ${\displaystyle \theta }$. Now, a magnetic needle lying parallel to a magnetic field requires W units of work to tum it through 60°. What will be the value of torque needed to maintain the needle in same position?

विकल्प

• ${\displaystyle \sqrt{3}}$ W

• W

• 0

• ${\displaystyle (\sqrt{3}/2)}$ W

Answer 1

${\displaystyle \sqrt{3}}$ W

Explanation:

Work done here is equivalent to magnetic potential energy of dipole.

${\displaystyle \therefore \text{W}=-\text{mB cos}\theta =-\text{mB}(\text{cos}{\theta }_2-\text{cos}{\theta }_1)}$

${\displaystyle =-\text{mB}(\text{cos} 60°)-(\text{cos} 0°)}$

${\displaystyle \text{W}=-\text{mB}(\frac{1}{2}-1)=\frac{\text{mB}}{2}}$   .... (i)

Now, when ${\displaystyle \theta }$ = 60°, torque acting oo dipole should be

${\displaystyle \tau =\text{mB sin}\theta =\text{mB sin} 60°=\frac{\sqrt{3}}{2}\text{mB}}$

Using (i)

${\displaystyle \tau =\sqrt{3} \text{W}}$