Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle θ with the field as shown in figure. If -m- is the strength of each pole, the two poles N and S experience two equal and opposite forces -mB-. These forces constitute a couple which tends to rotate the dipole. Suppose the couple exerts a torque of magnitude `tau.`

`tau = ` (either force) `times` (perpendicular distance between the two forces)


= mB `times` AC = mB `times` `2l sin θ`

or `tau= MB sin θ`

Here, `M = 2ml =` magnetic moment of the dipole

If B = 1 and θ = 90-

`tau= M`

Magnetic moment M of the magnetic dipole is defined as the moment of the couple acting on it when it is placed at right angles to a uniform field of unit strength.

In such case

Force on each pole `= m times 1`

Perpendicular distance between two forces `= NS = 2l`

So, `tau= m times 2l`

Thus, magnetic moment of a magnetic dipole is also defined as the product of its pole strength and the magnetic length.

Magnetic moment of a magnet is a vector quantity and is directed along SN inside the magnet.

`vecM = Mhatl`

`hat{l} =` unit vector along SN.

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

`vectau = vecMxxvecB`

The direction of `vec{tau }` can be obtained by applying right hand thumb rule.


`text(Units of M :-)`

Units of M can be obtained from the relation

`tau = MB sin θ`

So, `M = tau//B sin θ`

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

Since ,1 tesla `= (Wb)/m^2`

Thus, Unit of `M = J/((Wb)/m^2) = (Jm^2)/(Wb)`

Since 1 J = 1 Nm

Therefore, Unit of `M = 1 Nm T^(-1) = (1 Nm)/((Wb)/m^2) = 1 Nm^3 Wb^(-1)`

Since unit of pole strength = Am

Thus, Unit of `M = (Am) m = Am^2`


`text(Work done in rotating a magnetic dipole in a magnetic field)`

When a magnetic dipole of magnetic moment M is oriented in a magnetic field of strength, B, making an angle -θ-. With its lines of force, it experiences a torque `tau` given by

`vec{tau } = vec{M} times vec{B}`

Or

`|vec{tau } | = MB sintheta`

Let `dvectheta` be the small angular displacement given to the dipole, work done dW is given by

`dW = vectau. vec(d theta) = tau.d theta`

Since the angle between `vec{tau }` and `dvectheta` is zero.

So, `dW = MB sin θ dθ`

Work done in displacing the dipole from an angle `θ_1` to `θ_2` is

`W = int_0^W dW = int_(theta_1)^(theta_2) MB sintheta d theta`

`W = MB(costheta_1 - costheta_2)`


`text(Potential energy of a magnetic dipole in a magnetic field :)`

Potential energy of a magnetic dipole, in a magnetic field, is defined as the amount of work done in rotating the dipole from zero potential energy position to any desired position.
It is convenient to choose the zero potential energy position to be the one when the dipole is at right angles to the lines of force of magnetic field.

So, `θ_1 = 90-` and `θ_2 = θ`

Making these substitution in equation (3), we get,

Potential energy `= W = MB (cos 90- - cos θ)`

Or, `W = - MB cosθ`

In vector form, `W = - vec{M}.vec{B}`