A uniform straight magnetized wire has a magnetic moment M . If it...

Let the pole strength of the magnetized wire be $m$ and its length be $L$.

The initial magnetic moment is given by:
$M = m \times L$

When the wire is bent into an arc of a circle of radius $R$ subtending an angle $\theta = 90^{\circ} = \dfrac{\pi}{2}$ radians at the centre, the length of the arc remains $L$.

$L = R \theta = R \left(\dfrac{\pi}{2}\right)$

$\Rightarrow R = \dfrac{2L}{\pi}$

The new magnetic moment $M'$ depends on the shortest distance (chord length) $d$ between the two poles.

From the geometry of the circle, the chord length for an angle of $90^{\circ}$ is:
$d = \sqrt{R^2 + R^2} = \sqrt{2}R$

Substituting the value of $R$:
$d = \sqrt{2} \left(\dfrac{2L}{\pi}\right) = \dfrac{2\sqrt{2}L}{\pi}$

The new magnetic moment is:
$M' = m \times d = m \left(\dfrac{2\sqrt{2}L}{\pi}\right)$

$M' = \dfrac{2\sqrt{2}}{\pi} (mL) = \dfrac{2\sqrt{2}M}{\pi}$

Answer: $\dfrac{2\sqrt{2}M}{\pi}$