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# Derive an expression for the torque on a current carrying loop kept in a uniform magnetic field

Apoorva Arora IIT Roorkee
askIITians Faculty 181 Points
7 years ago
Consider a rectangular loop carrying a current I in the presence of a uniform magnetic field directed parallel to the plane of the loop, as shown in Figure. No magnetic forces act on sides 1 and 3 because these wires are parallel to the field; hence,LxB= 0 for these sides. However, magnetic forces do act on sides 2 and 4 because these sides are oriented perpendicular to the field. The magnitude of these forces is,
$F_{2}=F_{4}=IaB$
The direction ofF2, the force exerted on wire 2 is out of the page in the view shown in Figure, and that ofF4, the force exerted on wire 4, is into the page in the same view. If we view the loop from side 3 and sight along sides 2 and 4, we see the view shown in Figure, and the two forcesF2andF4are directed as shown. Note that the two forces point in opposite directions but are not directed along the same line of action. If the loop is pivoted so that it can rotate about point O, these two forces produce about O a torque that rotates the loop clockwise. The magnitude of this torque τmaxis
$\tau _{max}=F_{2}\frac{b}{2}+F_{4}\frac{b}{2}=IaB\frac{b}{2}+IaB\frac{b}{2}=IabB$
where the moment arm about O is b/2 for each force. Because the area enclosed by the loop is A = ab, we can express the maximum torque as
$\tau _{max}=IAB$
Thanks and Regards
Apoorva Arora
IIT Roorkee