Magnetic Dipole Moment is the property of a magnet that represents the magnetic strength and direction of the magnet. A magnetic dipole moment is also referred to as a magnetic moment of dipole and it is made up of two magnetic north poles separated by a short distance. The magnetic dipole moment is calculated as the product of the current and the area of the current carrying loop. Its unit is Ampere-(metre)2. Let's learn more about Magnetic Moment, Magnetic Dipole, Formula, Examples, and other related topics to dipole moment in this article.
What is Magnetic Moment?
Magnetic Moment is defined as the property which is used to define the strength and the orientation of the magnet. the magnetic moment is defined for fundamental charged particles, current-carrying loops, permanent magnets, and other things.
The magnetic moment of any particle is determined by using the magnetic dipole of the particle. The magnetic moment is a vector quantity and it allows the magnetic particle to arrange parallel to the external field if allowed to align freely in the magnetic field.
Magnetic Dipole
Magnetic Dipole also called magnetic dipole moment is defined as the vector quantity which tells us about the magnetic strength and orientation of any magnet. The magnetic dipole moment arranges the magnet in such a way that it is always parallel to the magnetic field and arranges itself in a North-South direction.
Magnetic Dipole Formula
The torque acting on the magnetic dipole is calculated by taking the vector product of the magnetic moment with the external magnetic field the formula for the same is,
τ = m × B
Where,
- τ is the torque acting on the magnetic dipole,
- m is the magnetic moment, and
- B is the strength of the external magnetic field.
Magnetic Moment Unit
As we know the magnetic moment is the product of the current and area then the SI unit of the magnetic moment is Am2, where A is the ampere and m is the meter. It can also be measured in Joule per Tesla or JT-1.
Derivation of Magnetic Dipole Moment Formula
The derivation to find the magnetic field strength due to magnetic dipole is discussed below.
Suppose the magnetic field due to a current loop carrying a current 'i' and the radius 'R' at a point P which is at a distance l along the axis is 'B' then the value of B is given using the formula,
B = µo iR2 / 2(R2 + l2)3/2
If the point P is very far from the current carrying loop i.e. R<<<l then the above formula is written as,
B = µo iR2 / 2(R2)3/2
B = µo iR2 / 2R3
Now rearranging the above formula we get,
B = µo/4π × 2i(πR2)/R3
Now we know that the area of the current carrying loop is A = πR2. Substituting this in the above formula we get,
B = µo/4π × 2iA/R3
Now if we define a new quantity magnetic as µ = iA the above formula becomes as,
B = µo/4π × 2µ/R3
This is very similar to the electric field formula of the electric dipole.
E = 1/4πεo × 2p/R3
The magnetic field of a magnet is described in the following image.
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 atomic model. An electric current is made up of a rotating electron in a confined channel. The anticlockwise movement of the electron generates a conventional current in the clockwise direction.
The current flowing through the conductor is given by the formula,
i = e ⁄ T...(1)
where
e is the charge
T is the time taken by the electron to complete one revolution
If v is the orbital velocity of the electron then,
T = 2πr ⁄ v...(2)
where r is the radius of the orbit.
Therefore from eq (1) and (2)
i = ev ⁄ 2πr...(3)
Suppose the orbital magnetic moment due to the electron's orbital motion is μl
μl = iA...(4)
where A is the area of the orbit then from eq. (3) and (4)
A = πr2
μl = (ev ⁄ 2πr) (πr2)
μl = evr ⁄ 2
Consider the mass of the electron to be m then,
μl = (e ⁄ 2m) × (mvr)
The angular momentum of electron (l) is given as l = mvr
μl = (e ⁄ 2m) × l
μl ⁄ l = e ⁄ 2m
Note: e ⁄ 2m is called the gyromagnetic ratio and is a constant and its value is 8.8 × 1010 C kg-1.
The angular momentum, according to Bohr.
l = nh ⁄ 2π
Where,
- n is a natural number
- h is Planck's constant
Substitute the formula of l into the formula of μl
μl = (e ⁄ 2m) × (nh ⁄ 2π)
Thus, μl = neh ⁄ 4πm
The minimum value of the magnetic moment is obtained when n = 1
μB = e h ⁄ 4 π m
OR
μB = e ħ ⁄ 2m
Where,
- ħ is reduced planck constant ħ = h/2π,
- m is the mass of electron at rest, and
- e is the charge on the electron.
Here, μB is called Bohr magneton that can be found by inserting the values of e, h, and m we get the value of the Bohr magneton to be 9.27 × 10–24 Am2.
The magnetic moment of the electron is shown in the image added below,
How Does an Atom Behave as Magnetic Dipole?
Electrons revolving around the nucleus revolve in closed orbits these revolving electrons produce the magnetic field and behave as magnetic dipoles.
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.
What is the Magnetic Dipole Moment of a Current Loop?
Suppose a current of strength 'I' passes through a circular loop of area 'A' and has 'n' turns around the loop. Then the magnetic dipole moment is calculated by using the formula,
µ = niA
Where,
- n is the number of turns the loop has made,
- i the current passing through the loop, and
- A is the area of the loop
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Solved Examples on Magnetic Dipole Moment
Example 1: What is the orbital magnetic moment of an electron moving with an orbital velocity of 0.5 m⁄s?
Solution:
Charge of the electron, e = 1.60217662 × 10−19 C
Radius of the electron, r = 2.817 940 3262 × 10−15 m
Velocity of electron = 0.5 m ⁄ s
Orbital magnetic moment, μl = e v r ⁄ 2
⇒ μl = 1.60217662 × 10−19 × 0.5 × 2.817 940 3262 × 10−15 ⁄ 2 A m2
⇒ μl = 1.1287095268 × 10−34 A m2
Hence, the orbital magnetic moment of an electron is 1.1287095268 × 10−34 A m2.
Example 2: If a current carrying loop having current 5 A has a cross-sectional area of 20 cm2 and the number of turns is 50 then find the magnetic dipole moment of the current carrying loop.
Solution:
Given,
- Current in the Circular loop (i) = 5 A
- Number of turns of the Loop (n) = 50
- Cross-Sectional Area of Loop = 50 cm2 = 0. 05 m2
Now the formula used is,
µ = niA
⇒ µ = 50×5×0.05
⇒ µ = 12.5 Am2
Thus, the magnetic moment of the current carrying loop is, 12.5 Am2.
Example 3: If a current-carrying loop has a magnetic moment of 25 Am2 and the cross-sectional area of the loop is 100 cm2 with the number of turns in the loop to be 5 then fins the current passing through the conductor.
Solution:
Given,
- Magnetic Moment of loop = 25 Am2
- Number of turns of the Loop (n) = 5
- Cross-Sectional Area of Loop = 100 cm2 = 0.1 m2
Now the formula used is,
µ = niA
⇒ 25 = 5×A×0.1
⇒ A = 50 A
Thus, the magnetic moment of the current carrying loop is, 50 A