Answer

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**Hint**So here in this question we have a steady current and their magnetic dipole moment is also given here. It is saying that if the loop of squares gets changed to a circular loop then it will carry the same current then we have to find the magnitude. So here we will use the concept which is of magnetic dipole and through this, we will get the relation.

Formula:

Magnetic dipole,

$m = nIA$ ;

Where,

$m$ , will be the magnetic dipole.

$n$ , will be equal to the turn in the loop and $I$ will be current and $A$ will be the area of current-conducting conductor

**Complete Step by Step Solution**Here we will first see the formula and accordingly we will put the given values and will get the result.

So from the formula, know the magnetic dipole is equal to the

$ \Rightarrow m = nIA$

Or we can write it as

$ \Rightarrow m = 1 \times I \times {a^2}$

Now we know from the question,

$ \Rightarrow 4a = 2\pi r$

Now we will solve the above equation for the value of $r$ , we get

$ \Rightarrow r = \dfrac{{2a}}{\pi }$

And as we already have seen the formula for the loop which is being circular, so

$ \Rightarrow m' = 1 \times I \times \pi {r^2}$

On substituting the values in the above equation, we will get

$ \Rightarrow 1 \times I \times \pi \times {\left( {\dfrac{{2a}}{\pi }} \right)^2}$

So from here, after solving the above equation, we will get the value for the magnitude of the magnetic dipole.

$ \Rightarrow m' = \dfrac{{4m}}{\pi }$

**Therefore, $\dfrac{{4m}}{\pi }$ is the required magnitude, and hence the correct option is B .**

**Note**A magnetic dipole is akin to an electric dipole, it exists in reality and has physical representation and meaning. The magnetic moment is just a consequence of mathematical building for physical proof. So we can say that when the field is produced because the magnetic field is proportional to the magnetic moment.

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