Magnetic Field due to a Short Bar Magnet | Magnetic Dipole - Class 12 physics Notes
Magnetic Field due to a Short Bar Magnet explains the magnetic field expressions at axial, equatorial, and arbitrary points around a magnetic dipole, highlighting their magnitudes, directions, and key relationships used in Class 12 Physics and JEE preparation.
Table of Contents
eSaral › Class 12 ›Magnetic Field due to a Short Bar Magnet
🚀 Checkout eSaral Courses
The magnetic field due to a short bar magnet at any point on the axial line is twice the magnetic field at a point on the equatorial line of that magnet at the same distance. S.l. unit of torque acting on the bar magnet is Nm.
Magnetic field due to a short bar magnet (magnetic dipole) :
On Axial Point or End on Position
The magnetic field $\overrightarrow{ B }_{\text {axial }}$ at a point P due to bar magnet will be the resultant of the magnetic fields $\overrightarrow{ B _{1}}$ due to N-pole of magnet and $\overrightarrow{ B _{2}}$ due to S-pole of magnet. 
$\overrightarrow{ B }_{\text {axial }}=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$
$\overrightarrow{ B }_{1}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r -\ell)^{2}}(\hat{ r })$ and $\overrightarrow{ B }_{2}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r +\ell)^{2}}(-\hat{ r })$
$\therefore \quad \vec{B}_{\text {axal }}=\left[\frac{\mu_{0}}{4 \pi} \frac{m}{(r-\ell)^{2}}-\frac{\mu_{0}}{4 \pi} \frac{m}{(r+\ell)^{2}}\right]$
$\hat{\mathrm{r}}=\frac{\mu_{0} \mathrm{~m}}{4 \pi}\left[\frac{4 \mathrm{r} \ell}{(\mathrm{r}-\ell)^{2}(\mathrm{r}+\ell)^{2}}\right] \hat{\mathrm{r}}$
$\overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ Mr }}{\left( r ^{2}-\ell^{2}\right)^{2}}$
Magnetic field due to a bar magnet at an axial point has same direction as that of its magnetic dipole moment vector.
For a bar magnet of very small length $\ell<< r \overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ M }}{r^{3}}$
Browse More Topics Related Magnetism:
On Equatorial Point or Broadside Position
The magnetic field $\overrightarrow{ B }_{\text {equi }}$ at a point P due to bar magnet will be the resultant of the magnetic fields $\overrightarrow{ B _{1}}$ due to N-pole of magnet and $\overrightarrow{ B _{2}}$ due to S-pole of magnet $\overrightarrow{ B }_{ equi }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$ due to S-pole of magnet $\overrightarrow{ B }_{ equ i }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$
$\left|\overrightarrow{ B }_{1}\right|=\frac{\mu_{0}}{4 \pi} \frac{ m }{ NP ^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ m }{ r ^{2}+\ell^{2}}$ along $NP$
$\left|\overrightarrow{ B }_{2}\right|=\frac{\mu_{0}}{4 \pi} \frac{ m }{ SP ^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)}$ along $PS$ 
So $\left|\vec{B}_{1}\right|=\left|\vec{B}_{2}\right|$ On resolving $\vec{B}_{1}$ and $\overrightarrow{ B _{2}}$ along PX' and PY we find $\left|\vec{B}_{1}\right|$ $\sin \theta$ and $|\overrightarrow {{B_2}} |\,\sin \theta $
are equal and opposite so they cancel each other. So resultant field
$\overrightarrow{\mathrm{B}}_{\text {equi }}=\overrightarrow{\mathrm{B}}_{1} \cos \theta(-\hat{\mathrm{r}})+\overrightarrow{\mathrm{B}}_{2} \cos \theta(-\hat{\mathrm{r}})$
$=\left[\frac{\mu_{0}}{4 \pi} \frac{m}{\left(r^{2}+\ell^{2}\right)} \cos \theta+\frac{\mu_{0}}{4 \pi} \frac{m}{\left(r^{2}+\ell^{2}\right)} \cos \theta\right](-\hat{r})$
$=2 \cdot \frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cdot \frac{\ell}{\sqrt{ r ^{2}+\ell^{2}}}(-\hat{ r })$
$\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$
$\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$
The direction of magnetic field at a point on equitorial line is opposite to magnetic dipole moment vector.
For a bar magnet of very small length $\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{r^{3}}$
At An Arbitrary Point
The point P is on axial line of magnet S'N' with magnetic moment Mcos$\theta $ Magnetic flux density $B _{1}=\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{ r ^{3}}$
The point P is simultaneously on the equatorial line of other magnet N"S" with magnetic moment Msin $\theta $ Magnetic flux density $B _{2}=\frac{\mu_{0}}{4 \pi} \frac{ M \sin \theta}{ r ^{3}}$
Total magnetic flux density at P.
$B =\sqrt{ B _{1}^{2}+ B _{2}^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{4 \cos ^{2} \theta+\sin ^{2} \theta}$ or
$\tan \theta^{\prime}=\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{r^{3}}}{\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}}}=\frac{1}{2} \tan \theta$ or $\theta^{\prime}=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)$ 
Also Read:
Click here for the Video tutorials of Magnetic Effect of Current Class 12
About eSaral At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.
For free video lectures and complete study material, Download eSaral APP.
Frequently Asked Questions
Find answers to common questions.
What is the magnetic field at the equatorial point of a short bar magnet?
The magnetic field at the equatorial point is B = (μ₀/4π)(M/r³), directed opposite to the magnetic dipole moment M. It is exactly half the axial field at the same distance. The general formula without the short-magnet approximation is B = μ₀M / [4π(r² + ℓ²)^(3/2)].
Why is the axial field twice the equatorial field?
For a short bar magnet at the same distance r, the axial formula gives 2M/r³ and the equatorial formula gives M/r³. The factor of 2 arises because at the axial point both poles contribute fields in the same direction, whereas at the equatorial point the perpendicular components cancel, halving the net result.
What is the formula for the magnetic field at an arbitrary point?
At a point P at distance r and angle θ from the magnetic axis, the field magnitude is B = (μ₀M/4πr³)√(3cos²θ + 1). The direction makes angle θ′ = tan⁻¹(tanθ/2) with the radial direction toward P. Substituting θ = 0° gives the axial result and θ = 90° gives the equatorial result.
What is the SI unit of magnetic dipole moment and torque on a bar magnet?
The SI unit of magnetic dipole moment (M) is A·m² (ampere metre squared). The SI unit of torque acting on a bar magnet placed in a uniform magnetic field is N·m (newton metre), same as the SI unit of work. The torque formula is τ = MB sinθ.
What is the direction of the magnetic field at the equatorial point compared to M?
The magnetic field at the equatorial point is antiparallel (opposite in direction) to the magnetic dipole moment vector M. This is a critical sign distinction: the axial field is parallel to M, while the equatorial field is antiparallel. Many board and competitive exam questions specifically test this directional knowledge.