A current carrying a small loop behaves like a small magnet. If A be its area and M its magnetic moment, the current in the loop will be
A. $\dfrac{W}{A}$
B. $\dfrac{A}{M}$
C. $\dfrac{M}{A}$
D. $A{M^2}$

Hint:We shall use the little loop carrying electricity to express its magnetic moment. In order to further clarify this notion, we can remark that the moment of a magnet is the measurement of the torque that the magnetic field will apply to it as well as the force that the magnet is capable of exerting on electric currents.



Formula Used:

The expression for the magnetic moment is available.
$M = IA$


Complete answer:

The quantitative representation of an object's tendency to fall in line with a magnetic field is known as a magnetic moment, sometimes known as a magnetic dipole moment. In other terms, a magnet or other item that generates a magnetic field's magnetic strength and orientation is referred to as the magnetic moment.
The product of the loop's magnetic moment M and the area A of the loop is equal to the loop's current I flowing through it.
We have provided A tiny loop that carries electricity that acts similarly to a small magnet with magnetic moment M. Let I be the current flowing through the loop.
The expression for the magnetic moment is available.
$M = IA$

Here, I is the current and A is the area.
The SI unit of magnetic dipole moment is given by $A{m^2}$
$I = \dfrac{M}{A}$
Hence option C is correct.



Thus, the correct option is C.



Note: According to the right-hand thumb rule, the magnetic dipole moment (M) is a vector field whose direction is perpendicular to A,$M = IA$. The thumbs in these rules indicate the direction of the magnetic field, and the grip indicates the direction of the current flow. Two processes—motion electric charge and spin angular momentum—produce the magnetic moment.