**Magnetic Moment Formula**

Magnetic Moment is an extremely important topic in the syllabus of Physics in JEE Advanced. The magnetic moment of a magnet can be defined as the quantity that finds the force a magnet is able to exert on electric currents and the torque that the magnetic field will exert on it. Magnetic moments can be demonstrated by a lot of things such as a loop of electric current, an electron, a bar magnet, a molecule or a planet. In this article, find all about the magnetic moment formula.

Magnetic moment can be basically explained as a vector that has magnitude as well as a direction. The direction of the magnetic moment points from the south to the North Pole of a magnet. Please note that the magnetic field created by a magnet is directly proportional to its magnetic moment. The term magnetic moment is about the system’s magnetic dipole moment. The dipole component of an object’s magnetic field is symmetric about the direction of its magnetic dipole moment and decreases as the inverse cube of the distance from the object.

Magnetic moment is also known as a magnetic dipole moment and its positive direction is dependent on the way an object reacts to the magnetic field. Objects tend to place themselves in such a way that the magnetic moment vector then becomes parallel to the magnetic field lines. Magnetic moment can be produced in two ways:

- The Motion of Electric Charge
- Spin Angular Momentum

For instance, a loop of wire that has a current running through would be having a magnetic moment, which is proportional to the current and also the area of the loop points in the direction of your right thumb if your fingers are curling in the direction of the current. Read on to find out more magnetic moment formula.

**Magnetic Dipoles**

A magnetic dipole primarily decided the limit of either a current loop or a pair of poles as the dimensions of the source get reduced to zero, while the moment is kept as constant. When these limits get applied to fields that are far away from sources, they remain the same. But note that these two models are quite different for the internal field. Magnetic dipole moment in physics for IIT JEE is quite an important topic for all the JEE applicants as it makes the foundation for various other topics strong.

When we consider a magnetic dipole as a current loop, the magnitude of the dipole moment is proportional to the current multiplied by the size of the enclosed area. The direction of the dipole moment can also be represented mathematically as a vector and is perpendicularly away from the side of the surface enclosed by the counterclockwise path of positive charge flow.

When we consider the current loop to be a tiny magnet, this vector corresponds to the direction from the south to the North Pole. Dipoles manage to align themselves so that their moments point largely in the direction of the external magnetic field when they are free to rotate. Nuclear and electron magnetic moments can be quantified and that implies that they may be oriented in space at only certain discrete angles with respect to the direction of the external field.

Magnetic dipole moments have dimensions of current time’s area or energy divided by magnetic flux density. The exact unit for dipole moment is ampere-square metre when it comes to the metre–kilogram– second–ampere and SI systems. But when it comes to the centimetre–gram–second electromagnetic system, the unit is the erg (unit of energy) per gauss (unit of magnetic flux density). Around one thousand ergs per gauss are equal to one ampere-square metre. The Bohr magneton (equivalent to 9.27 × 10−24 ampere–square metre) is quite a fitting unit for the magnetic dipole moment of electrons. A similar unit for magnetic moments of nuclei, protons, and neutrons is the nuclear magneton (equivalent to 5.051 × 10−27 ampere–square metre).

**External Magnetic Field produced by a Magnetic Dipole Moment **

If there’s a system that includes a net magnetic dipole moment m, then it produces a dipolar magnetic field. However, the net magnetic field created by the system may display comparatively higher-order multipole components. But it’s interesting to know that they fall out quite rapidly with distance and in the end, just the dipolar component dominates the magnetic field of the system at distances far away from it.

**Internal Magnetic Field of a Dipole**

It’s important to know that both the models for a dipole provide the same result for the magnetic field far from the source. However, all the predictions, which happen inside the source region, are different. The magnetic field between poles is in the opposite direction to the magnetic moment.

**Magnetic Moment Formula: Calculation for a Bar Magnet**

(a) If a magnet of length l and magnetic moment M gets bent in the form of a semicircular, then its new magnetic moment will be M’ = 2M/π

(b) The magnetic moment of a given electron because of its orbital motion is 1μ_{B}. But due to its spin motion, it will be μ_{B }/2.

i.e. M_{orbital} =

and M_{spin} = s

Here μ_{B} = Bohr magneton

(i) The value of Bohr magneton μ_{B} = eh / 4πm

(ii) μB = 0.93 × 10^{–23} Amp-m^{2}

(c) **Other formulae for magnetic moments:**

(i) M = ni r^{2}

(ii) M = eVr/2 = er^{2}ω/2 = er^{2}2πf/2 = er^{2}π/T

(iii) e/mJ

(iv) M = nω_{B}

(d) **How can we calculate resultant magnetic moment? **

(i) When two bar magnets are lying mutually perpendicular to each other, then

M = √M_{1}^{2} + M_{2}^{2} = √2m_{p}l

(ii) When two coils with a radius of r and carrying current I each, are lying concentrically with their planes at right angles to each other, then

M = √M_{1}^{2} + M_{2}^{2} = √2iπr^{2} if M_{1} = M_{2}

**Illustration: A square loop OABCO of side l carries a current i. Now, find the magnetic moment of the loop.**

* Solution: *Magnetic moment of the loop can be written as,

**M **= i (**BC** x **CO**), where the letters in bold denote the vectors

**BC** = -l**k**

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