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Derive an expression for the torque expe...

Derive an expression for the torque experienced by a magnetic dipole in a uniform magnetic field and hence define magnetic dipole moment.

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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.
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