Derive an expression for the torque acting on a current carrying loop suspended in a uniform magnetic field.

Explanation

1. As shown in figure, consider a rectangular coil ABCD suspended in a uniform magnetic field `vecB`, with its axis perpendicular to the field.


2. Suppose, current flowing through coil ABCD be I, AB = DC = b and AD = BC = a and area be A = ab.
2. The field exerts no force on the two arms AD and BC of the loop, because `I_(a)andvecB` both are parallel to each other,
`|vecF|=|IvecaxxvecB|`
`|vecF|=(IaxxB)sin0^(@)=0" "...(1)`
3. Sides AB and CD of coils are perpendicular to magnetic field.
4. So, force on side AB is,
`vecF_(1)=|IvecbxxvecB|=Ib"sin"pi/2`
`|vecF_(1)|=IbB" "...(2)`
5. Similarly it exerts a force `vecF_(2)` on the arm CD and `vecF_(2)` is directed out of the plane of the paper,
`|vecF_(2)|=|IvecbxxvecB|=|Ib"sin"pi/2|`
`thereforeF_(2)=IbB=F_(1)" "...(2)`
6. Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces `vecF_(1)andvecF_(2)`.
7. Figure (b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anti-clockwise.
8. Resultant torque on coil is given by,
`tau=tau_(1)+tau_(2)`
`tau=F_(1)(a/2)+F_(2)(a/2)`
= `IbB(a/2)+IbB(a/2)`
= `I(ab)B`
`tau=IAB" "...(3)`
where A = ab that is area of rectangle.