Question

A circular coil of $100$ turns and an effective diameter of $20cm$ carries a current of $0.5A$ . It is to be turned in a magnetic field $2.0\;T$ from a position in which the normal to the plane of the coil makes an angle $\theta$ equal to zero to one in which $\theta$ equals $180$ . The work required in this process is:
(A) $\pi J$
(B) $2\pi J$
(C) $4\pi J$
(D) $8\pi J$

Answer 1

We know the current is produced due to the flow of electrons. A current-carrying wire produces a magnetic field. Also, when a current-carrying wire is placed in a magnetic field it experiences a magnetic force. The direction of force is found by Fleming’s right-hand rule.

Complete answer:
Work done is stored as the potential energy of the system. So, we have the following formula.
$W = U = - \vec m.\vec B$
Here, $U$ is potential energy, $m$ is magnetic dipole and $B$ is a magnetic field.
Again, the magnetic dipole is calculated as below.
\vec m = IA\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{n}
Here, $I$ is current, $A$ is an area, and $\hat n$ is the unit normal.
Let us first write the information given in the question.
$N = 100$ , $d = 20cm \Rightarrow r = 10cm$ , $I = 0.5A$ , $B = 2T$ , ${\theta _1} = 0$ , ${\theta _2} = 180$
Magnetic moment: $m = \left( {0.5} \right)\left( {\pi {{\left( {.10} \right)}^2}} \right) = 0.005\pi$
Work done will be a change in the potential energies.
$W = mB\cos {\theta _1} - mB\cos {\theta _2}$
We can rewrite this expression as below.
$W = mB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$
Let us substitute the values.
$W = 0.005\pi \times 2\left( {\cos 0 - \cos 180} \right)$
Let us simplify the above expression.
$W = 0.01\pi \left( {1 - \left( { - 1} \right)} \right) = 0.02\pi J$
Now, it is given that there are $100$ turns. So total work done will become.
$W = 100 \times 0.02\pi = 2\pi J$
Therefore, total work done is $2\pi J$ .
Hence, option (B) $2\pi J$ is correct.

Note:
The magnetic dipole moment of an object is defined as the torque experienced by the object in a magnetic field. Unit of a magnetic dipole is $A - {m^2}$ .