Question

A circular coil of radius 4 cm and 20 turns carries a current of 3 amperes. It is placed in a magnetic field of intensity of 0.5 weber/$m^2$ .The magnetic dipole moment of the coil is
(A) 0.15 ampere-$m^2$
(B) 0.3 ampere-$m^2$
(C) 0.45 ampere-$m^2$
(D) 0.6 ampere-$m^2$

Answer 1

Hint: - We know that magnetic dipole moment of coil will depend upon the area of the coil, the number of turns and the current in the coil, but we have also been given the magnetic field intensity of the field in which the coil is kept, when clearly it has no effect on the magnetic dipole moment of the coil, so we’ll need to keep an eye out for such deceptions from the examiner.

Step by Step Solution:
Given that $radius(r)=4cm=0.04m$
Number of turns, $n=20$
Current in the coil, $i=3A$
We know that, area of a circular coil, $A=\pi {{r}^{2}}$
Therefore area of our current carrying coil, $A=\pi \times {{(0.04)}^{2}}=0.00502{{m}^{2}}\simeq 0.0050{{m}^{2}}$(approximating)
We also know that magnetic dipole moment, $M=niA=20 \times 3 \times 0.0050=0.3A-{{m}^{2}}$

Note: Here the solution is pretty basic but they have given an extra quantity to try and lure you into confusion.
You can also be sure of the method by observing the unit of all of the options as sometimes that might help you choose the correct option as well. Here the unit is ampere-$m^2$. We know that ampere is the unit of current and $m^2$ is the unit of area, so we need to take the product of the area of the coil, no. of turns of coil(which has no unit), and the current in the coil. The unit of magnetic field intensity is weber-$m^2$ but that doesn’t concern us here because it doesn’t appear in any of the options. This explanation is only to double check your answer. You should follow the step by step procedure to obtain the answer.