Magnetic Dipole Moment

The magnetic dipole moment is a key physical quantity that describes the magnetic strength and direction of a system, such as a current loop, bar magnet, or electron. It determines how the system interacts with an external magnetic field, experiencing torque that aligns it with the field and storing potential energy based on its orientation. In current loops, it is proportional to the current and loop area, while at the microscopic level, it arises from the orbital motion and spin of charged particles. This concept is fundamental in understanding magnetism and its applications in physics and engineering.

1.0Magnetic Dipole

• A magnetic dipole is an arrangement of two equal and opposite magnetic poles separated by a small distance.

• Unlike electricity, where the simplest structure is an isolated point charge, isolated magnetic poles (magnetic monopoles) do not exist in nature. The simplest magnetic structure is always the dipole.
• A magnetic dipole is described by a vector quantity called the magnetic dipole moment (m).
• Bar Magnet: Every bar magnet is a classic example of a magnetic dipole. Current-Carrying Loop: A current-carrying loop of wire behaves as a magnetic dipole.
• Atoms: Even an atom acts as a magnetic dipole. This is caused by the circulatory motion of the electrons around its nucleus

2.0Magnetic Dipole Moment

• Its direction is taken from the South Pole to the North Pole.

$\vec{m}=q_m\times 2\vec{l}$

• It is a vector quantity.
• Its direction is taken from the South pole to the North pole.
• The SI unit of magnetic dipole moment is ampere metre squared $\left(A{m}^2\right)$ or joule per tesla $\left(J{T}^{-1}\right)$.

3.0Axial Magnetic Field of A Bar Magnet

Consider a unit north pole is situated at point,force exerted by north pole is given as

Force due to the north pole:

$F_N=\frac{{\mu }_0}{4\pi }\frac{q_m}{(r-l{)}^2}$ along NP

Force exerted by south pole is given as

$F_S=\frac{{\mu }_0}{4\pi }\frac{q_m}{(r+l{)}^2}$ along PS

Magnetic field at axial point:

$B_{\text{axial}}=F_N-F_S$

$B_{\text{axial}}=\frac{{\mu }_0}{4\pi }{q}_m\left(\frac{1}{(r-l{)}^2}-\frac{1}{(r+l{)}^2}\right)$

$B_{\text{axial}}=\frac{{\mu }_0}{4\pi }\frac{4q_{m}rl}{(r^2-l^2{)}^2}$

$m=q_m\cdot 2l$

$B_{\text{axial}}=\frac{{\mu }_0}{4\pi }\frac{2mr}{(r^2-l^2{)}^2}$

For a short magnet $((l\ll r))$

: ${\vec{B}}_{\text{axial }}=\frac{{\mu }_0}{4\pi }\frac{2\vec{m}}{r^3}$

4.0Equatorial Magnetic Field of a Bar Magnet

Consider a unit north pole is situated at point,force exerted by north pole is given as

$F_N=\frac{{\mu }_0}{4\pi }\cdot \frac{q_m}{x^2}\text{ along }\overrightarrow{NP}$

Force exerted by south pole is given as

$F_S=\frac{{\mu }_0}{4\pi }\cdot \frac{q_m}{x^2}\text{ along }\overrightarrow{PS}$

Magnetic Field at the equatorial point P is

$B_{\text{Equatorial }}=F_N\cos \theta +F_S\cos \theta$

$B_{\text{Equatorial }}=2F_N\cos \theta \left[\because F_N=F_S\right]$

$B_{\text{Equatorial }}=2\frac{{\mu }_0}{4\pi }\cdot \frac{q_m}{x^2}\cdot \frac{l}{x}\left[\because \cos \theta =\frac{l}{x}\right]$

$B_{\text{Equatorial }}=\frac{{\mu }_0}{4\pi }\cdot \frac{m}{{\left(r^2+l^2\right)}^{3/2}}\left[\because x={\left(r^2+l^2\right)}^{1/2}\right]$

$m=q_m\cdot 2l$

For a short bar magnet (l<<r)

$B_{Equa}=\frac{{\mu }_0}{4\pi }\frac{m}{r^3}$

${\vec{B}}_{Equa}=\frac{{\mu }_0}{4\pi }\frac{\vec{m}}{r^3}$

5.0Magnetic Moment of Current Carrying Coil

The current carrying coil (or loop) behaves like a magnetic dipole. The face of the coil in which current appears to flow anticlockwise acts as north pole while the face of the coil in which current appears to flow clockwise acts as south pole. A current carrying coil acts as a Magnetic Dipole.

• For a coil of 'N', geometrical Area 'A', carries a current 'I'

Magnetic Moment, $\vec{M}=NI\vec{A}$

Where $\vec{A}$ is the area vector perpendicular to the plane of the coil along its axis SI Unit $A-m^2$ or $J/T$

• The direction of $\vec{M}$ can be found out by the right-hand thumb rule
• Curling fingers - In the direction of the current.
• Thumb - Gives the direction of $\vec{M}$.
• For a current carrying coil, its magnetic moment and magnetic field vectors both are parallel axial vectors.

6.0Magnetic Dipole Moment of a Revolving Electron

• Based on the Bohr model for hydrogen-like atoms, the magnetic dipole moment $(\mu )$ of a revolving electron.
• A negatively charged electron orbits a positively charged nucleus in uniform circular motion. This motion is equivalent to a current loop.
• A current loop possesses a magnetic dipole moment $(\mu )$
• The magnetic dipole moment $(\mu )$ is defined by the product of the current (I) and the area (A) of the loop, $\mu =I.A$
• Consider an electron of charge (e) revolving anticlockwise in an orbit of radius r with speed v and time period T.

$I=\frac{e}{T}=\frac{ev}{2\pi r}$

Area of the current loop $,A==\pi r^2$

Orbital Magnetic Moment

${\mu }_l=IA=\frac{ev}{2\pi r}\cdot \pi r^2$

${\mu }_l=\frac{evr}{2}\dots \dots (1)$

Angular momentum of electron due to its orbital motion, $l=m_{e}vr\dots (2)$

Using equation (1) and (2)

$\frac{{\mu }_l}{l}=\frac{\frac{\epsilon vr}{2}}{m_{\epsilon }vr}=\frac{e}{2m_{\epsilon }}=\text{ Gyromagnetic Ratio }=8.8\times 10^{10}CK{g}^{-1}$

${\mu }_l=\frac{e}{2m_{\epsilon }}l$

$\overrightarrow{{\mu }_l}=-\frac{e}{2m_{\epsilon }}\vec{l}$