NS is the bar magnet of length 2l and pole strength m. P is a point on the equatorial line at a distance d from its mid point O (Fig.).

**Magnetic induction at a point along
the equatorial line of a bar magnet**

NS
is the bar magnet of length 2*l* and
pole strength m. P is a point on the equatorial line at a distance d from its
mid point O (Fig.).

Magnetic
induction (B_{1}) at P due to north pole of the magnet,

B_{1}
= ?_{0}/4π . m/NP^{2} along NP

=
?_{0}/4π . m/(d^{2}+*l*^{2}) along NP

NP^{2}
= NO^{2} + OP^{2}

Magnetic
induction (B_{2}) at P due to south pole of the magnet,

B_{2}
= ?_{0}/4π . m/PS^{2} along PS

=
?_{0}/4π . m/(d^{2}+*l*^{2}) along PS

Resolving
B1 and B2 into their horizontal and vertical components.

Vertical
components B1 sin θ and B2 sin θ are equal and opposite and therefore cancel
each other (Fig.).

The
horizontal components B1 cos θ and B2 cos θ will get added along PT.

Resultant
magnetic induction at P due to the bar magnet is

B
= B_{1} cos θ + B_{2} cos θ. (along PT)

After
apply B_{1} and B_{2}

B
= = ?_{0}/4π . M/d^{3}

The
direction of ?B? is along PT parallel to NS.

*Coulomb?s inverse square law*

Coulomb?s inverse square law
states that the *force of attraction or*
*repulsion between the two magnetic poles
is directly proportional to the product of their pole strengths and inversely
proportional to the square of the distance between them.*

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