Derive an expression for torque acting on a bar magnet placed in a uniform magnetic field.

Explanation

Consider a rectangular loop ABCD carrying current I in anticlockwise direction i.e., in direction ABCD.
Let `l = AB = CD = ` Length of the rectangular loop.
`b = BC = DA = ` Breadth of the rectangular loop.
`vecB=` Strength of magnetic field directed from left to right in the plane of paper.
`theta =` Angle between the normal to plane of coil and direction of magnetic field.
Force acting on the arm ABof the loop.
`vecF_(1) = I(vecl xx vecB)`

Direction of `vecF_(1)` is perpendicular to the length of arm AB and directed outward of the sheet of paper (Fleming.s Left Hand Rule ).
Similary, force acting on the arm CD of the loop.
`vecF_(2) = I(vecl xx vecB) " "...(ii)`
According to Fleming.s Left Hand Rule, `vecF_(2)`is normal to the length of arm CD and is directed inside the sheet of the paper.
Forces acting on arm BC and on arm DA are equal and opposite to each other and act alongthe same line of action and hence cancel out in pairs.
`vecF_(1) and vecF_(2)` are also equal and opposite to each other, but their line of action is not same.
So `vecF_(1) and vecF_(2)` form a couple and try to rotate the loop anticlockwise.
The magnitude of torque is given by
`tau = ` Either force `xx`Perpendicular distance between `vecF_(1) and vecF_(2)`.
`tau = F_(1) xxDN`
`tau = I(vecl xx vecB) xx DN`
` = I(lBsin 90^(@)) xx b sin theta`
` = I(lb) B sin theta`
`tau = IAB sin theta`
where `A = lb =` Area of the loop
if the loop has N turns then
`tau = NIABsin theta`
This torque will be maximum when `theta = 90^(@)`
So `tau_("maxi") = NIAB`
This torque will be minimum when `theta = 0^(@)`
So `tau_("mini") = NIAB sin 0`
`tau_("mini") =0`