The magnetic field to a small magnetic dipole of magnetic moment $M$ , at distance $r$ from the centre on the equatorial line is given by (in M.K.S. system)
A. $\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{r^2}}}$
B. $\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{r^3}}}$
C. $\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2M}}{{{r^2}}}$
D. $\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2M}}{{{r^3}}}$

Hint:
Magnetic dipole is an arrangement of two different magnetic poles of equal strength separated by a small distance. The magnetic field due to a small magnet at any point on the equatorial line is half of the magnetic field at a point on the axial line of that magnet at the same distance.

Formula used:
The formula used in the solution of this problem is: -
Magnetic field on equatorial line, ${B_{equitorial}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{{\left( {{r^2} + {l^2}} \right)}^{\dfrac{3}{2}}}}}$
Magnetic field on axial line, ${B_{axial}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2Mr}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}}$



Complete step by step solution:
We know that the magnetic field from the centre on the equatorial line of a bar magnet is given by: -
${B_{equitorial}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{{\left( {{r^2} + {l^2}} \right)}^{\dfrac{3}{2}}}}}$ … (1)
And the magnetic field from the centre on the axial line of a bar magnet is given by: -
${B_{axial}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2Mr}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}}$ … (2)
where $M$ = magnetic dipole moment
and $l$ = length of a magnetic dipole.
Since the length of a magnetic dipole is very small according to the question i.e., $l < < < r$therefore, $l$ is negligible.
From eq. (1)
${B_{equitorial}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{{\left( {{r^2}} \right)}^{\dfrac{3}{2}}}}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{r^3}}}$
and, from eq. (2)
 ${B_{axial}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2Mr}}{{{{\left( {{r^2}} \right)}^2}}} = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2M}}{{{r^3}}}$
Also, we can write as: - ${B_{axial}} = 2{B_{equitorial}}$
Thus, the magnetic field on the equatorial line is given by $\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{r^3}}}$ in M.K.S. system.
Hence, the correct option is (B) $\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{M}{{{r^3}}}$ .



Therefore, the correct option is B.




Note:
The component of the magnetic moment which can be represented by a magnetic dipole. Magnetic dipole is a magnetic analogue of electric dipole. Only one exception is there. a magnetic monopole which is a magnetic analogue of electric charge are not found in nature.