Magnetic dipole moment of a current loop

The magnetic dipole moment is $M = I A$ where $I$ is the current and $A$ is the area enclosed by the loop.

For the square loop

Let the side of the square be $a$.

Then its area is $A_s = a^2$

Given that its magnetic dipole moment is $m$, so $m = I a^2$

When the square is changed into a circular loop

Since the same wire is used, the perimeter remains the same.

Perimeter of square: $4a$

Circumference of circle: $2\pi r$

Therefore, $2\pi r = 4a$ $r = \frac{2a}{\pi}$

Area of the circular loop

$A_c = \pi r^2 = \pi \left(\frac{2a}{\pi}\right)^2$

$A_c = \pi \cdot \frac{4a^2}{\pi^2} = \frac{4a^2}{\pi}$

Magnetic dipole moment of the circular loop

$M_c = I A_c = I \cdot \frac{4a^2}{\pi}$

Using $m = Ia^2$, $M_c = \frac{4m}{\pi}$

Checking options

• A: $\dfrac{m}{\pi}$
• B: $\dfrac{3m}{\pi}$
• C: $\dfrac{2m}{\pi}$
• D: $\dfrac{4m}{\pi}$

Hence the correct option is $\boxed{\frac{4m}{\pi}}$ i.e. Option D.