A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

The magnetic dipole moment of any planar current-carrying loop is given by the relation

$$m = NI A,$$

where $$N$$ is the number of turns, $$I$$ is the current and $$A$$ is the area enclosed by the loop. In both the square and the circular loop we have a single turn, so $$N = 1$$ throughout.

For the original square loop let the length of each side be $$a$$. Its area is therefore

$$A_{\text{square}} = a^{2}.$$

Because it carries the current $$I$$, its magnetic dipole moment is given to be $$m$$, so from the defining formula we can write

$$m = I \, a^{2}.$$

Solving this expression for $$a^{2}$$ gives

$$a^{2} = \frac{m}{I}.$$

The same wire is now bent into a circle, so its length does not change. The perimeter of the square equals the circumference of the circle. Thus

$$4a = 2\pi r,$$

where $$r$$ is the radius of the circle. Dividing both sides by $$2\pi$$ we obtain

$$r = \frac{2a}{\pi}.$$

Next we compute the area of the circle:

$$A_{\text{circle}} = \pi r^{2} = \pi\left(\frac{2a}{\pi}\right)^{2}.$$

Squaring the numerator and denominator and then cancelling one factor of $$\pi$$, we get

$$A_{\text{circle}} = \pi \cdot \frac{4a^{2}}{\pi^{2}} = \frac{4a^{2}}{\pi}.$$

Now we substitute $$a^{2} = \dfrac{m}{I}$$, obtained earlier, into this area:

$$A_{\text{circle}} = \frac{4}{\pi}\left(\frac{m}{I}\right).$$

The magnetic dipole moment of the circular loop, denoted $$m'$$, is again $$NI A$$ with $$N = 1$$, so

$$m' = I \, A_{\text{circle}} = I \left(\frac{4}{\pi}\frac{m}{I}\right).$$

Here the current $$I$$ in the numerator cancels the $$I$$ in the denominator, leaving

$$m' = \frac{4m}{\pi}.$$

Thus, after reshaping the square loop into a circular loop while keeping the same current, the magnetic dipole moment becomes $$\dfrac{4m}{\pi}$$.

Hence, the correct answer is Option A.