Derive an expression for the magnetic field intensity at a point on the axial line of a bar magnet (magnetic dipole). Also, state the expression for a short bar magnet.

Q: Derive an expression for the magnetic field intensity at a point on the axial line of a bar magnet (magnetic dipole). Also, state the expression for a short bar magnet.

Introduction\nA bar magnet behaves as a magnetic dipole consisting of two equal and opposite magnetic poles separated by a small distance. To find the magnetic field at an axial point, we consider the contribution of both the North pole and the South pole. Derivation Let a bar magnet of length $2l$ and pole strength $m$ have its magnetic dipole moment given by $M = m \times 2l$. Let $P$ be a point on the axial line at a distance $r$ from the center of the magnet ($O$). The distance of point $P$ from the North pole ($N$) is $(r - l)$, and from the South pole ($S$) is $(r + l)$. The magnetic field at $P$ due to the North pole ($BN$) is directed away from $N$ and is given by: $$BN = \frac{\mu0}{4\pi} \frac{m}{(r-l)^2}$$ The magnetic field at $P$ due to the South pole ($BS$) is directed towards $S$ and is given by: $$BS = \frac{\mu0}{4\pi} \frac{m}{(r+l)^2}$$ The net magnetic field $B$ at point $P$ acts along the direction of $BN$ (since $N$ is closer than $S$): $$B = BN - BS$$ $$B = \frac{\mu0}{4\pi} m \left \frac{1}{(r-l)^2} - \frac{1}{(r+l)^2} \right$$ $$B = \frac{\mu0}{4\pi} m \left \frac{(r+l)^2 - (r-l)^2}{(r^2 - l^2)^2} \right$$ $$B = \frac{\mu0}{4\pi} m \left \frac{4rl}{(r^2 - l^2)^2} \right$$ Substituting $M = m \times 2l$: $$B = \frac{\mu0}{4\pi} \frac{2Mr}{(r^2 - l^2)^2}$$ For a Short Bar Magnet\nIf the magnet is very short compared to the distance $r$ ($l \ll r$), then $l^2$ can be neglected in comparison to $r^2$: $$B = \frac{\mu0}{4\pi} \frac{2Mr}{r^4} = \frac{\mu0}{4\pi} \frac{2M}{r^3}$$

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Accepted Answer

Introduction
A bar magnet behaves as a magnetic dipole consisting of two equal and opposite magnetic poles separated by a small distance. To find the magnetic field at an axial point, we consider the contribution of both the North pole and the South pole. Derivation Let a bar magnet of length $2l$ and pole strength $m$ have its magnetic dipole moment given by $M = m 	imes 2l$. Let $P$ be a point on the axial line at a distance $r$ from the center of the magnet ($O$). The distance of point $P$ from the North pole ($N$) is $(r - l)$, and from the South pole ($S$) is $(r + l)$. The magnetic field at $P$ due to the North pole ($BN$) is directed away from $N$ and is given by: $$BN = \frac{\mu0}{4\pi} \frac{m}{(r-l)^2}$$ The magnetic field at $P$ due to the South pole ($BS$) is directed towards $S$ and is given by: $$BS = \frac{\mu0}{4\pi} \frac{m}{(r+l)^2}$$ The net magnetic field $B$ at point $P$ acts along the direction of $BN$ (since $N$ is closer than $S$): $$B = BN - BS$$ $$B = \frac{\mu0}{4\pi} m \left \frac{1}{(r-l)^2} - \frac{1}{(r+l)^2} ight$$ $$B = \frac{\mu0}{4\pi} m \left \frac{(r+l)^2 - (r-l)^2}{(r^2 - l^2)^2} ight$$ $$B = \frac{\mu0}{4\pi} m \left \frac{4rl}{(r^2 - l^2)^2} ight$$ Substituting $M = m 	imes 2l$: $$B = \frac{\mu0}{4\pi} \frac{2Mr}{(r^2 - l^2)^2}$$ For a Short Bar Magnet
If the magnet is very short compared to the distance $r$ ($l \ll r$), then $l^2$ can be neglected in comparison to $r^2$: $$B = \frac{\mu0}{4\pi} \frac{2Mr}{r^4} = \frac{\mu0}{4\pi} \frac{2M}{r^3}$$

CBSE · Class 12 · Physics · Magnetism and MatterDerive an expression for the magnetic field intensity at a point on the axial line of a bar magnet (magnetic dipole). Also, state the expression for a short bar magnet.

Introduction\nA bar magnet behaves as a magnetic dipole consisting of two equal and opposite magnetic poles separated by a small distance. To find the magnetic field at an axial point, we consider the contribution of both the North pole and the South pole.

Derivation

For a Short Bar Magnet\nIf the magnet is very short compared to the distance $r$ ($l \ll r$), then $l^2$ can be neglected in comparison to $r^2$: