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A small coil of N turns has area A and a current I flow through it. The magnetic dipole moment of this coil will be
(A) $iNA$
(B) ${i^2}NA$
(C) $i{N^2}A$
(D) $IN/A$

Answer
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163.2k+ views
Hint: Start with finding what the magnetic dipole moment and how it works in general. Then apply it in the case of the coil with a certain number of turns, having area and current flowing through it. After doing you will get the required answer and there will be no exact value as it will be in form of variables.

Complete answer:
First start with what is a magnetic dipole moment.
In general, magnetic dipole moment is the magnetic strength and the orientation(direction) of a magnet or the other object that produces a magnetic field. It is a vector quantity and is related to the current loops and magnets (in general). Now in case of coil that has number of turns, area and current flowing through it.

Magnetic dipole moment will be equal to product of current flowing through the coil and the area of that coil.
So,
$\vec m = IA$
Where,
$\vec m = magnetic\,dipole\,moment$
I is the current in the coil
A is area of the coil
Also the coil has number of turns N.
Therefore, $\vec m = NIA$

Hence the correct answer is Option(A).

Note: Use the basic knowledge of the magnetic dipole moment carefully then put the values provided in the question in the general formula for the magnetic dipole moment and finally you will get the required answer in terms of the number of turns, area of the coil and current flowing through the coil.