# Magnetic Dipole Moment of a Revolving Electron

A magnetic moment is a measurement of a magnet’s magnetic strength and direction, as well as any other item that creates a magnetic field. A magnetic moment is more accurately referred to as a magnetic dipole moment, which is the component of the magnetic moment that can be represented by a magnetic dipole. A magnetic dipole is made up of two magnetic north poles separated by a short distance.

The dimensions of magnetic dipole moments are current times area or energy divided by magnetic flux density.

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Ampere-square meter is the unit for dipole moment in meter–kilogram–second–ampere. The erg (unit of energy) per gauss is the unit in the centimeter–gram–second electromagnetic system (unit of magnetic flux density). One ampere-square meter equals 1,000 ergs per gauss.

### Magnetic Dipole Moment of a Revolving Electron

Most elementary particles are magnetic dipoles by nature. The electron, for example, possesses a Spin Magnetic Dipole moment and acts as a magnetic dipole. This magnetic moment is essential to the nature of the electron’s existence, as the electron has neither an area A (it is a point object) nor does it rotate about itself.

The negatively charged electron revolves around a positively charged nucleus in a circular orbit of radius r, according to Neil Bohr’s atom model. An electric current is made up of a rotating electron in a confined channel. The anticlockwise travel of the electron generates a conventional current in the clockwise direction.

Current is given as:

**i = e ⁄ T**

where, e is the charge of an electron and T is the period of the electron’s revolution.

If v is the electron’s orbital velocity, then:

**T = 2 π r ⁄ v**

where, r is the radius of the orbit.

Therefore,

**i = e v ⁄ 2 π r**

There will be an orbital magnetic moment μ_{l} due to the electron’s orbital motion.

**μ _{l} = i A**

Here, A is the area of the orbit.

**A = π r ^{2}**

**μ _{l} = (e v ⁄ 2 π r) (π r^{2})**

**μ _{l} = e v r ⁄ 2**

Consider the mass of the electron be m.

**μ _{l} = (e ⁄ 2 m) (m v r)**

The electron’s angular momentum (l) around the central nucleus is (m v r).

**μ _{l} = (e ⁄ 2 m)l**

(μ_{l} ⁄ l = e ⁄ 2 m is called **gyromagnetic ratio** and is a constant.)

Its value is 8.8 × 10^{10} C kg^{-1}. The angular momentum, according to Bohr, has only a discrete set of values given by the equation.

**l = n h ⁄ 2 π**

Here,

n is a natural number and h is Planck’s constant.

Substitute the formula of l into the formula of μ_{l}.

**μ _{l} = (e ⁄ 2 m) (n h ⁄ 2 π)**

** = n e h ⁄ 4 π m**

The minimum value of the magnetic moment is obtained at **n = 1**.

**μ _{l} = e h ⁄ 4 π m**

It is called **Bohr magneton**.

The value of the Bohr magneton is determined to be 9.27 × 10^{–24} Am^{2} by inserting the values of e, h, and m.

The electron has a magnetic moment due to its spin in addition to the magnetic moment due to its orbital motion. As a result, an electron’s resultant magnetic moment is the vector sum of its orbital and spin magnetic moments.

### Sample Questions

**Question 1: What is the behaviour of an atom as a magnetic dipole?**

**Answer:**

In an atom, electrons are in a confined orbit around the nucleus. Because electrons are charged particles, their orbit around the nucleus is analogous to a current loop. The electrons spin in an anticlockwise direction, whereas the current spins in a clockwise direction. The flow of electrons results in the formation of a south pole and a north pole, causing the atom to behave like a magnetic dipole.

**Question 2: What is a Current Loop’s Magnetic Dipole Moment?**

**Answer:**

The magnitude of m is the magnetic dipole moment of a current loop carrying current i with area A.

m = i A

The magnetic dipole moment has a direction that is perpendicular to the plane of the current loop.

**Question 3: What is the value of current flowing in the electron for a time period of 2 s?**

**Answer:**

Charge of the electron, e = 1.60217662 × 10

^{−19}CGiven:

Time period, T = 2 s

Current, i = e ⁄ T

= 1.60217662 × 10

^{−19}⁄ 2 A= 0.80108831 A

Hence, the current flowing in the electron is

0.80108831 A.

**Question 4: What is the orbital magnetic moment of an electron moving with an orbital velocity of 0.5 m ⁄ s?**

**Answer:**

Charge of the electron, e = 1.60217662 × 10

^{−19}CRadius of the electron, r = 2.817 940 3262 × 10

^{−15}mGiven that,

Velocity of electron = 0.5 m ⁄ s

Orbital magnetic moment, μ

_{l}= e v r ⁄ 2= 1.60217662 × 10

^{−19}× 0.5 × 2.817 940 3262 × 10^{−15}⁄ 2 A m^{2}= 1.1287095268 × 10

^{−34}A m^{2}Hence, the orbital magnetic moment of an electron is

1.1287095268 × 10.^{−34}A m^{2}

**Question 5: Why do electrons have a magnetic field?**

**Answer:**

The movement of electric charges causes magnetism. Each atom contains electrons, which are charged particles. Electrons spin like tops around the nucleus, or centre, of an atom. Each electron acts as a tiny magnet as a result of their mobility, which creates an electric current.