Question

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 \[{180^0}\] will be:
A. \[ - MB\]
B. \[ + MB\]
C. \[ + 2MB\]
D. \[Zero\]

Answer 1

Hint:
To solve this question we have to use the basic formula of work done in moving a dipole in an external magnetic field. Use the given data and by putting it into the equation we can directly solve the question.

Formula used:
\[W = MB(1 - \cos \theta )\]
\[M\]- magnetic moment of the dipole
\[B\]- magnetic field strength


Complete step by step solution:
let us solve the given question by using the given data.
Given data: \[M\]- magnetic moment of the dipole.
             \[B\]- magnetic field strength
             \[\theta = {180^0}\]
Now, by using formula for work done in moving the dipole with magnetic moment in given magnetic field by the angle of \[{180^0}\] we have:
\[W = MB(1 - \cos \theta )\]
By using \[\theta = {180^0}\]in above equation we get
\[ \Rightarrow W = MB(1 - \cos {180^0})\]
\[ \Rightarrow W = MB(1 - ( - 1))\]
\[ \Rightarrow W = 2MB\]
Hence, the work done on the magnet to rotate it with the angle of \[{180^0}\]is \[2MB\].
Correct answer is option c.




Therefore, the correct option is C.




Note:
In this question, the dipole is along the magnetic field also called stable equilibrium position and rotating the magnet by \[{180^0}\] then it will be at unstable equilibrium position. A particle always tries to remain at a stable equilibrium position so whenever we move the dipole from its stable equilibrium position we have to do some extra work.