The correct option is A √7μ0M2πd3
Since both M and 3M are making 120∘, therefore the resultant magnetic moment is,
Mres=√M21+M22+2M1M2cosθ
=√(M)2+(3M)2+2M(3M)cos120∘
=√M2+9M2+6M2(−12)
=√7M2=√7 M.
Now point P is on the axial line of the resultant magnetic moment.
By symmetry, it is clear that the resultant magnetic field is along the perpendicular bisector.
∴B=μ04π(2Md3)
B=μ04π(2√7 Md3)
B=√7 μ0M2πd3
Hence, option (A) is the correct answer.