A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane. 

EXPLANATION:

→ Magnetic moment of a current carrying loop M = IA 

Where I = Current and A = Area of the loop = πR² (Where, R = radius of the circular loop)

∴ M = IπR2 ---- (1)

The direction of the magnetic moment is given by right hand-thumb rule (as shown)

⇒Half of the loop with x > 0 is now bent so that it lies in the y-z plane as shown in the figure below.

The magnitudes of magnetic moments of each semi-circular loop of radius R lying the xy plane and yz plane 

⇒ M₁ = M₂ = I(A/2)

= ½ IπR² =  ½ M  [ from (1) ]

The directions of M1 and M2 are along the z-axis and x-axis respectively (as shown).

The resultant, M_net  = $\sqrt{M_1^2+M_2^2}=\sqrt{2\times \frac{1}{4}\times }M=\frac{M}{\sqrt{2}}$

So, Mnet  < M 

∴ M diminishes. 

So, the correct answer is option (1).

ref. image for the dtp team