Question

Derive the expression for magnetic field due to a bar magnet at an arbitrary point.​

Answer 1

Answer:

i) When the point is on the axis of the magnet: $\\\tt B_{net}= \dfrac{ \mu }{4 \pi} \times \dfrac{2 M_{moment} }{a^{3} }$

iI) When the point is on the perpendicular to the axis of the magnet: $\\\tt B_{net} = \dfrac{\mu}{4 \pi} \times \dfrac{M_{moment}}{l^{3 }} }$

Explanation:

i) Axial

Let a magnet with poles separated by 2l distance with pole strength M and a point a units away form the center  of the magnet,

Now,

$\\\tt B_{north \ pole} = \dfrac{ \mu }{4 \pi} \times \dfrac{M}{(a-l)^{2} } \\ \\\tt B_{south \ pole} = - \dfrac{ \mu }{4 \pi} \times \dfrac{M}{(a+l)^{2} } \\ \\\tt B_{net} = B_{north \ pole} + B_{south \ pole} \\ \\\tt B_{net} = \dfrac{ \mu M }{4 \pi} \times [ \dfrac{1}{(a-l)^{2} }+ \dfrac{1}{(a+l)^{2} }] \\ \\\tt = \dfrac{ \mu M }{4 \pi} \times \dfrac{4al}{(a^{2}- l^{2})^{2} }$

Since a>>l, we have

$\\\tt B_{net}= \dfrac{ \mu M }{4 \pi} \times \dfrac{4l}{a^{3} }$

Since, 2lM= magnetic moment,

$\\\tt B_{net}= \dfrac{ \mu }{4 \pi} \times \dfrac{2 M_{moment} }{a^{3} }$

ii) Equatorial position

$\\\tt B_{net} =2 \times \dfrac{\mu}{4 \pi} \times \dfrac{M cos(\theta)}{( \sqrt{ l^{2}+ a^{2}} )^{2}}$

Since cos θ = $\\\tt \dfrac{l}{( \sqrt{ l^{2}+ a^{2}} )}$

Using this,

$\\\tt B_{net} = \dfrac{\mu}{4 \pi} \times \dfrac{2Ml}{( l^{2}+ a^{2})^{\dfrac{3}{2} }} }$

Since once again, a>>l and 2Ml = magnetic moment,

$\\\tt B_{net} = \dfrac{\mu}{4 \pi} \times \dfrac{M_{moment}}{l^{3 }} }$