Question

derive an expression for the magnetic induction due to short magnetic dipole at a point on its equator​

Answer 1

Answers :-

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

.

Explanation :-

Answer 2

Answer:

Explanation:

Consider a bar magnet NS of length 2l and pole strength m.

Let P be the point on the equatorial line at distance d' from mid -point

of magnet .

Now , the magnet induction ,$B_{1}$ at point due to north pole along NP is ,

$B_{1} = \frac{\mu_{0} m}{4\pi NP^{2} }$

$B_{1}$ = $\frac{\mu_{0} m}{4\pi (d^{2} +l^{2} )^{2} }$ ..........(i)   ($NP^{2} = ON^{2}+OP^{2}$)

Similarly , the magnetic induction ,$B_{2}$ at point P due tosouth ple along PS is ,$B_{2} = \frac{\mu_{0} m}{4\pi PS^{2} }$

⇒ $B_{2} = \frac{\mu_{0} m}{4\pi (d^{2} +l^{2} )^{2} }$ ..........(II)     ($PS^{2} = OS^{2}+OP^{2}$)

From equation (i) and (ii)

$B_{1} =B_{2}$= B (say)

Now, the net magnetic induction $B_{p}$ at P due to bar magnet is

$B_{p} = B_{1} +B_{2} = 2B .........(3)$

When resolving $B_{1}$ and $B_{2}$ into horizontal and vertical components .

The vertical components will be canceled and horizontal will be added together .

Therefore , $B_{p} = 2Bcos\theta$   (from equation iii )

$B_{p}$ = $2\frac{\mu_{0} m}{4\pi (d^{2} +l^{2} )^{2} }(\frac{1}{\sqrt{d^{2} +^{l2} } } )$       (where     $cos\theta = (\frac{1}{\sqrt{d^{2} +^{l2} } } )$)

For a short dipole ,l<<d, therefore

$B_{p} = \frac{\mu_{0} 2ml}{4\pi d^{3} }$  .

Final answer :

Hence , here we drived an expression for the magnetic induction due to short magnetic dipole at a point on its equator .