Derivation of magnetic field on equitorial line of dipole
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The shortest distance between the two poles of a bar magnet is the magnetic length of the bar magnet (2l). The axial line of a bar magnet is a straight line passing through its centre and its poles. The equatorial line of a bar magnet is a straight line perpendicular to its axial line and passing through its centre. The magnetic moment M of a magnet is defined as the product of its magnetic length 2l and its pole strength m. Therefore, M = 2lm. The SI unit of magnetic moment M is ampere metre-square (Am2) or weber metre (Wb.m). The magnetic moment of a bar magnet is a vector quantity whose direction is along its axial line, from its south pole towards its north pole.
The magnetic induction B at a point in the magnetic field is equal to the force experienced by a unit north pole placed at that point. If an isolated north pole of strength m placed at a point in a magnetic field experiences a force F due to the magnetic field, then magnetic induction B at that point in the magnetic field is given by
B = F/m ------ (1)
Magnetic induction at a point along the equatorial line of a bar magnet
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.).
Figure 1 See Upward
Magnetic induction (B1) at P due to north pole of the magnet,
B1 = μ0/4π . m/NP2 along NP
= μ0/4π . m/(d2+l2) along NP
NP2 = NO2 + OP2
Magnetic induction (B2) at P due to south pole of the magnet,
B2 = μ0/4π . m/PS2 along PS
= μ0/4π . m/(d2+l2) along PS
Figure 2 See Upward
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 = B1 cos θ + B2 cos θ. (along PT)
After apply B1 and B2
B = = μ0/4π . M/d3
The direction of ?B? is along PT parallel to NS.
On an axial line
B=μ0/4π*2Mr/(r^2-l^2)^2
If r<<1
B=μ0/4π*2M/r^3
The magnetic induction B at a point in the magnetic field is equal to the force experienced by a unit north pole placed at that point. If an isolated north pole of strength m placed at a point in a magnetic field experiences a force F due to the magnetic field, then magnetic induction B at that point in the magnetic field is given by
B = F/m ------ (1)
Magnetic induction at a point along the equatorial line of a bar magnet
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.).
Figure 1 See Upward
Magnetic induction (B1) at P due to north pole of the magnet,
B1 = μ0/4π . m/NP2 along NP
= μ0/4π . m/(d2+l2) along NP
NP2 = NO2 + OP2
Magnetic induction (B2) at P due to south pole of the magnet,
B2 = μ0/4π . m/PS2 along PS
= μ0/4π . m/(d2+l2) along PS
Figure 2 See Upward
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 = B1 cos θ + B2 cos θ. (along PT)
After apply B1 and B2
B = = μ0/4π . M/d3
The direction of ?B? is along PT parallel to NS.
On an axial line
B=μ0/4π*2Mr/(r^2-l^2)^2
If r<<1
B=μ0/4π*2M/r^3
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