Question

# Derive an expression for magnetic field intensity due to a magnetic dipole at a point lies on its equatorial line.

Solution

## Consider a point $$P$$ on equatorial (or broad side on position) of short bar magnet of length $$2I$$, having north pole $$(N)$$ and south pole $$(S)$$ of strength $$+q_m$$ and $$-q_m$$ respectively. The distance of point $$P$$ from the mid-point $$(O)$$ of magnet is $$r$$. Let $$B_1$$ and $$B_2$$ be the magnetic field intensities due to north and south poles respectively. $$NP=SP=\sqrt {r^2}+l^2$$. $$vec {B_1}=\dfrac {\mu_0}{4\pi}\dfrac {q_m}{r^2 +l^2}$$ along $$N$$ to $$P$$ $$vec {B_2}=\dfrac {\mu_0}{4\pi}\dfrac {q_m}{r^2 +l^2}$$ along $$P$$ to $$S$$ Clearly, magnitude of and are equali.e., $$|\vec {B_1}|=|\vec {B_2}|$$ or $$B_1 =B_2$$To find the resultant of $$\vec {B_1}$$ and $$\vec {B_2}$$ we resolve them along and perpendicular to magnetic axis $$SN$$ components $$\vec {B_1}$$ of along and perpendicular to magnetic axis are $${B_1}\cos \theta$$ and $${B_2}\sin \theta$$ respectively.Components of $$\vec {B_2}$$ along and perpendicular to magnetic axis are $$2 \cos \theta$$ and $${B2}\sin \theta$$ respectively. Clearly components of and perpendicular to axis $$SN$$. $${B1}\sin \theta$$ and $${B2}\sin \theta$$ ae equal in magnitude and opposite in direction and hence, cancel each other, while the components of $$\vec {B1}$$ and $$\vec {B2}$$ along the axis are in the same direction and hence, add up to give to resultant magnetic filed parallel to the direction $$\vec {NS}$$.$$\therefore \$$ Resultant magnetic field intensity at $$P$$.$$B=B_1 \cos \theta +B_2 \cos \theta$$But $$B_1=B_2 =\dfrac {\mu_0}{4\pi}\dfrac {q_m}{r^2+l^2}$$an$$\cos \theta =\dfrac {ON}{PN}=\dfrac {1}{\sqrt {r^2 +l^2}}=\dfrac {1}{(r^2 +l^2)^{1/2}}$$$$\therefore \ B=2B_1 \cos \theta =2\times \dfrac {\mu_0}{4\pi}\dfrac {q_m}{(r^2 +l^2)}\times \dfrac {l}{(r^2 +l^2)^{1/2}}=\dfrac {\mu_0}{4\pi}\dfrac {2q_m l}{(r^2 +l^2)^{3/2}}$$But $$q_m2I=m$$, magnetic moment of magnet$$\therefore \ B=\dfrac {\mu_0}{4\pi}\dfrac {m}{(r^2 +l^2)^{3/2}}$$If the magnet is very short and point $$P$$ is far away, we have $$I << r$$; so $$12$$ may be neglect as compared to $$r^2$$ and so equation (3) takes the form $$\boxed {B\dfrac {\mu_0}{4\pi}\dfrac {m}{r^3}}$$This is the expression for magnetic field intensity at the equatorial position of the magnet.It is clear from equation (2) and (4) that the magnetic field strength due to a short magnetic dipole is inversely proportional to the cube of its distance from the center of the dipole and the magnetic field intensity at axial position is twice that at the equatorial position for the same distance.Physics

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