## Introduction

We know that if an electric dipole is placed in a uniform electric field, it will feel some amount of torque; likewise, if a magnetic dipole is placed in a uniform magnetic field, a net torque will be acted on it. We represent magnetic flux density with ‘B’ and magnetic dipole moment with ‘M’.

Magnetic dipole moment can be defined for any object having magnetic property. We have also learnt this term in atomic sciences because atoms have some magnetic properties, too (discussed in later paragraphs). In the next few paragraphs, we will discuss how a dipole in a uniform external field acts, and the torque acting on a dipole and try to analyse them with the help of some mathematical expressions.

### Definition of magnetic dipole moment

Magnetic dipole moment is the tendency of an object to interact with the external magnetic field. In simple words, it is the measurement of an object’s magnetic strength. More dipole moment means a stronger magnet. It is a vector quantity and denoted with μ. The term magnetic dipole is not only valid for bar magnets but can also be defined for current-carrying loops. In different cases, the expression for dipole moment changes.

#### Magnetic dipole placed in a uniform external magnetic field: torque acting on a dipole

When a dipole is placed in a uniform magnetic field, it experiences a net torque. No force is applied if the field is uniform. That torque aligns the dipole along the magnetic field. If two individual and opposite magnetic poles are placed two close, then a dipole is formed (although practically magnetic monopoles do not exist).

Suppose ‘B’ is the magnetic field strength and ‘m’ is the magnetic pole strength , distance between the poles is 2l and be the torque, then the expression will be:

= (either force) x (perpendicular distance between the two forces)

Now , the force acting on each pole will be = mB (equal in magnitude opposite in direction) and if θ be the angle between the magnetic field and the dipole then the perpendicular distance between the two forces becomes = 2lsinθ.

So the torque will be

= mB x 2lsinθ

= 2lm x B x sinθ (M= 2lm= magnetic dipole moment)

Since M and B, both, are vectors the torque acting on a magnet suspended in the magnetic field can be expressed as the cross-product of M and B as follows.

= M B

#### Units of M

Units of M can be obtained from the relation:

= MB sin θ

So, M = /B sin θ

Therefore, M can be measured in terms of joule per tesla (JT-1) or (NmT-1).

Since ,1 tesla = Wb/m2

Thus, Unit of M = J/(Wb/m2) = Jm2/Wb

Since 1 J = 1 Nm

Therefore, Unit of M = 1 Nm T-1 = 1 Nm/(Wb/m2) = 1 Nm3 Wb-1

Since unit of pole strength = Am

Thus, Unit of M = (Am) m = Am2

### Work done in rotating a dipole in uniform magnetic field

If a dipole is placed in a uniform magnetic field making and angle θ with the field then the torque acts on that dipole is:

= M B

Let the dipole is given a small displacement d, then the work done dW is given by:

dW = . d = d

= MB sin θ

dW = MB sinθdθ

So the total work done in displacing the dipole from angle θ1 to θ2

W = 0WdW = 12MB sin d

W = MB [(-cos θ2) – (- cos θ1)]

Or, W = MB (cos θ1 – cos θ2)

### Potential energy stored in a dipole in a uniform magnetic field

If some amount of work is done against some force, it gets stored in that system in potential energy. The potential energy of a magnetic dipole in a magnetic field is defined as the amount of work done in rotating the dipole. Let the zero potential energy position be the one when the dipole is at right angles to the lines of force of the magnetic field.

So, θ1 = 90º and θ2 = θ

According to the above equation of work done:

Potential energy = W = MB (cos 90º – cos θ)

Or, W = – MB cosθ

In vector form, W = -M. B

#### Special cases

(a) if θ = 0º

So, W = – MBcos0

W = -MB

(b) If θ = 90º

So, W = – MBcos90

W = -MB(0)= 0

(c) When θ = 180º

So, W = -MBcos(180)

W = – MB (-1) = MB

### Current carrying loops and magnetic dipole moment

Magnetic field lines are also generated from current-carrying solenoid/ loops. Magnetic lines of similar types are also present in bar magnets. By carefully observing those two symmetries, Ampere stated that a current-carrying loop behaves like a bar magnet and shows properties like it. It is called Ampere’s hypothesis.

Now let’s deduce the expression for the dipole moment in a current-carrying loop.

The magnetic induction at a point along the axis of a circular coil carrying current is,

B = µ0nIa2/2(a2+x2)3/2

The direction of this magnetic field is along the axis and is given by right hand rule. If a point is very far then, x>>a, a2 is small and it is neglected. Hence for such points,

B = µ0nIa2/2×3

If we consider a circular loop, n = 1, its area A = πa2

So, B = µ0IA/2πx3

The magnetic induction at a point along the axial line of a short bar magnet is

B = (µ0/4π) (2M/x3)

Or, B = (µ0/2π) (M/x3)

Comparing the equations, we find that

M = IA

So a current loop can be treated as a dipole whose dipole moment, M = IA

For a current carrying loop it’s dipole moment is defined as the product of current and area and directed along the area vector A.

### Magnetic dipole moment in atoms

It is discussed in the introduction that atoms also have some dipole moment. But the question is why? The origin of the dipole moment in atoms is basically due to the presence of revolving electrons. Previously we have discussed how and why a current-carrying loop Shows dipole moment. Electron being a revolving charged particle also shows the same effect as a current-carrying loop. As a result, it also has some amount of magnetic dipole moment.

Since the electron is negatively charged, conventional current flows in a direction opposite to its motion. An electron revolving in an orbit of radius ‘r’ is equivalent to a magnetic shell of magnetic moment ‘M’ given by:

M = iA

Here ‘i’ is the current to which the rotating electron is equivalent to and A is the area of the orbit. If ‘’ is the time period of rotation of electrons

i = charge/period = e/ = e/(2π/ω)

or i = eω/2π

Here, ‘ω’ is the angular velocity of the electron.

So, M = (eω/2π) (πr2) = eωr2/2

According to Bohr’s theory, an electron can revolve in an orbit in which its angular momentum is an integral multiple of h/2π where ‘h’ is Planck’s constant.

mr2ω = n (h/2π)

Or, r2ω = n (h/2πm)

Substituting for r2ω we get,

M = n (eh/4πm)

Now applying the above equation we can find the value of magnetic dipoles for each electron.

As dipole moment is a vector quantity in case of multielectron species the value of net dipole moment of an electron is the vector sum of all the individual dipole moments.

### Conclusion

Although electric dipoles in a uniform electric field and bar magnet in the magnetic field have the same kind of behaviour, there are certainly other cases in magnetism that are not there in electrostatics. This concept is very important in research as well as in engineering fields. Spectroscopy and magnetic resonance tests are two of their important uses of it.