Question

Magnetic moment of bar magnet is $M$. The work done in turning the magnet by ${90^0}$ in direction of magnetic field $B$ will be
A. Zero
B. $\dfrac{1}{2}MB$
C. $3MB$
D. $MB$

Answer 1

Hint:The magnetic moment of a magnet is defined as the quantity that determines the torque it will exerted on in an external magnetic field. Magnetic moment is defined as a vector relating the ranging torque on the object from an externally applied magnetic field to the field vector itself.

Formula used:
$W = - MB\cos \theta $ is used where $W$ is the potential energy, $M$ is the magnetic moment of the bar magnet, $B$ is the strength of the magnetic field and $\theta $is the angle made by the magnetic moment with the magnetic field.

Complete step by step answer:
It is given that the Magnetic moment of the bar magnet is $M$ and the magnetic field strength is $B$ and it is given that the magnet is turned by ${90^0}$. The initial angle is equal to ${0^0}$. We know that the potential energy of a magnetic moment at an angle $\theta $ with the magnetic field is equal to $ - MB\cos \theta $. Thus the work done in rotating the magnet from an angle ${\theta _1}$ to an angle ${\theta _2}$ will be equal to
$
W = - MB(\cos {\theta _2} - \cos {\theta _1}) \\
W= MB(\cos {\theta _1} - \cos {\theta _2})$
Substituting the values of ${\theta _1}$ and ${\theta _2}$ in the equation $W = MB(\cos {\theta _1} - \cos {\theta _2})$, we get
$W = MB(\cos {0^0} - \cos {90^0})$
$ \Rightarrow W = MB(1 - 0)$
$ \therefore W = MB$

Hence,option (D) is the correct answer.

Note: If a magnet is placed in an external magnetic field $B$, it will experience a torque. The magnitude of the torque depends on the orientation of the magnet with respect to the magnetic field. Magnetic moments are shown by various things including an electric current loop, a bar magnet, an electron, a molecule or a planet.