Question

A small ball of volume V made of a paramagnetic object of magnetic susceptibility $\chi$was slowly displaced along the axis of a current carrying coil from the point where the magnetic field induction is B out to the region where the magnetically is practically absent. What amount of work was done in the process?

Answer 1

Hint: When we place a paramagnetic object in this case a ball of volume V, the object will get magnetized due to the external field B. Work has to be against this magnetic energy in order to move the body out of the magnetic field.

Complete step by step answer:
In this problem we are given a paramagnetic ball of volume V and magnetic susceptibility $\chi$ which is placed on the axis of a current carrying coil. So first of all, a paramagnetic object is an object that gets magnetized along the external magnetic field because they have magnetic dipoles present in them, which align along the external field when an external field is applied.
When a paramagnetic object is placed in an external field B, the energy stored in a small volume dV is given by,
$\text{U}=\dfrac{1}{2}\int{\overrightarrow{H\text{ }}\centerdot \overrightarrow{\text{ }B}dV}$ ………..equation (1)
Where,
H is the magnetic intensity of the paramagnetic object.
B is the external magnetic field.
dV is the small volume we are considering
We know that, $\text{B}=\mu \text{H}$, where $\mu$ is the absolute permeability of the medium. So substituting for H in equation (1), we will get
$\text{U}=\dfrac{1}{2}\int{\dfrac{\overrightarrow{B}}{2{{\mu }_{0}}(1+\chi )}\centerdot \overrightarrow{B}}\text{ }dV=\dfrac{{{B}^{2}}}{2{{\mu }_{0}}(1+\chi )}\int{dV}$
We can write absolute permeability as $\mu ={{\mu }_{0}}(1+\chi )$
If we integrate dV we will get the volume of the ball which is V. So the above equation can be written as,
$\text{U}=\dfrac{{{B}^{2}}V}{2{{\mu }_{0}}(1+\chi )}$ ……..equation (2)

The force applied on the paramagnetic ball by the magnetic field B is given by,
$\text{F}=-\dfrac{dU}{dx}$
Since in U, only the magnetic field changes along the axis, others are constant. So we can write,
$\text{F}=-\dfrac{V}{2{{\mu }_{0}}(1+\chi )}\dfrac{d{{B}^{2}}}{dx}$
So the force required to move the ball in the coil is the negative of this force. We can denote that force to move the ball as ${{\text{F}}_{\text{m}}}$.
${{F}_{m}}=-F$
So small amount of work that should be done to move the ball through a small distance dx is,
$dw={{F}_{m}}dx$
Let the point where the magnetic field is zero be f. So the work done in moving the ball from initial point i to a point f where the magnetic field is practically zero is,
$W=\int\limits_{i}^{f}{{{F}_{m}}}dx$
$\text{W}=\dfrac{V}{2{{\mu }_{0}}(1+\chi )}\int\limits_{i}^{f}{\dfrac{d{{B}^{2}}}{dx}}dx=\dfrac{V}{2{{\mu }_{0}}(1+\chi )}\int\limits_{i}^{f}{d{{B}^{2}}}$
The magnetic field at point ‘i’ is B, while the magnetic field at point ‘f’ is zero. Therefore, the total work done is,
$\text{W}=-\dfrac{{{B}^{2}}V}{2{{\mu }_{0}}(1+\chi )}$
The negative sign indicates that work is done against the magnetic field B.

Note: The susceptibility value $\left( \chi \right)$for paramagnetic materials is of the order ${{10}^{-3}}$.
The susceptibility value $\left( \chi \right)$for diamagnetic materials is less than zero.
The susceptibility value $\left( \chi \right)$for ferromagnetic materials is very much greater than one $\left( \chi >>1 \right)$