$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

  • A
    $2 \sqrt{3} W$
  • B
    $\sqrt{3} W$
  • C
    $\frac{\sqrt{3}}{2} W$
  • D
    $\frac{\sqrt{3}}{4} W$

Explore More

Similar Questions

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

Two identical magnetic dipoles of magnetic moments $1.0 \, A-m^2$ each are placed at a separation of $2 \, m$ with their axes perpendicular to each other. The resultant magnetic field at a point $P$ midway between the dipoles is:

Difficult
View Solution

$A$ magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength $B$. The work done in rotating it by an angle of $180^{\circ}$ will be

The work done in rotating a bar magnet,which is initially in the direction of a uniform magnetic field,through $45^{\circ}$ is $W$. The additional work to be done to rotate the magnet further through $15^{\circ}$ is

The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ is .........

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo