Question

A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle ${180^ \circ }$, then the work is
 (M= magnetic dipole moment of bar magnet)
A. $MB$
B. $2MB$
C. $\dfrac{{MB}}{2}$
D. Zero

Answer 1

Hint: The work done to rotate the bar magnet is given by the formula $W = MB\left( {1 - \cos \theta } \right)$, where M is the magnetic dipole moment, B is the magnetic field of induction. Substitute the value of the angle in this formula to find the work done by the bar magnet.

Complete step by step answer:
We are given that a bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B.
We have to calculate the work done by the bar magnet when it is rotated through an angle ${180^ \circ }$
We have to use the formula $W = MB\left( {1 - \cos \theta } \right)$ to calculate the work done by rotating a bar magnet with an angle θ.
$
  W = MB\left( {1 - \cos \theta } \right) \\
  \theta = {180^ \circ } \\
  \cos {180^ \circ } = - 1 \\
  \Rightarrow W = MB\left( {1 - \cos {{180}^ \circ }} \right) \\
  \Rightarrow W = MB\left( {1 - \left( { - 1} \right)} \right) \\
  \Rightarrow W = MB\left( {1 + 1} \right) \\
  \Rightarrow W = 2MB \\
 $
When a bar magnet is rotated through an angle of ${180^ \circ }$, then the work done will be $2MB$

Note:A bar magnet is a rectangular piece of an object made up of iron or steel that shows permanent magnetic properties. It has two poles, a north and a south pole such that when suspended freely, the magnet aligns itself so that the northern pole points towards the magnetic north pole of the earth. Magnetic Dipole Moment is described as the product of pole strength and the distance between the two poles. When all the forces that act upon an object are balanced, then the object is said to be in equilibrium.