Derive an expression for the torque acting on a bar magnet in the uniform magnetic field an hence define magnetic dipole moment.

Explanation

Magnetic dipole moment is defined as, the product of the pole strength and the distance between two poles. If m be the strength of each pole and 2l be the distance between two poles, then

`vecM = m(2vecl)`
Magnetic dipole moment is a vector quantity, directed from south to north pole.
Unit: The SI unit of `vecM " are joule/tesla " (J T^(-1))` or ampere `"metre"^2 (A m^2)`. Torque on bar magnet in magnetic field : Consider a bar magnet NS of length 2l. Let m be the strength of each pole. Let the bar magnet be placed in uniform magnetic field `vecB . " Let " theta` be the angle between axis of magnet and direction of `vecB`. Force on N pole = mB along the direction of `vecB`. Force on S pole = mB along the direction opposite to `vecB`.

These two forces are equal and opposite and they act at different points. Hence they form a couple which tends to rotate the magnet clockwise. Draw NA perpendicular on SA produced. Torque acting on bar magnet `(iota)` will be given by
`iota` = Force x perpendicular distance
`iota = mB xx NA` ......(i)
From `DeltaNSA, (NA)/(NS) = sin theta " " implies NA = sin theta = 2l sin theta`
Put in equation (i), we get `iota = mB xx 2l sin theta`
`iota = (m xx 2l ) B sin theta " " implies iota = MB sin theta`
In vector form, `veciota = vecM xx vecB` . The direction of `iota` is perpendicular to the plane containing `vecM and vecB` as shown in figure. If `B = 1 " unit, " theta = 90^@, " then " iota = Mxx 1 xx sin 90^@ implies iota = M` Therefore, magnetic dipole moment is the torque acting on a magnetic dipole placed normal to a uniform magnetic field of unit strength.