A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is [2016]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is [2016]

1 $\frac{W}{\sqrt{3}}$

2 $\sqrt{3} W$

3 $\frac{\sqrt{3} W}{2}$

4 $\frac{2 W}{\sqrt{3}}$

Ans.
(2)

At equilibrium, potential energy of dipole

$U_{i} = - M B_{H}$

Final potential energy of dipole, $U_{f} = - M B_{H} \cos 60 ° = - \frac{M B_{H}}{2}$

$W = U_{f} - U_{i} = - \frac{M B_{H}}{2} - (- M B_{H}) = \frac{M B_{H}}{2}$ ...(i)

$τ = M B_{H} \sin 60 ° = 2 W \times \frac{\sqrt{3}}{2} [Using eqn. (i)]$

$= \sqrt{3} W$

Want Us to Email you About Special Offers & Updates?