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Question

The figure shows a circular wire loop of radius a and carrying a current i, which is placed in a perpendicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field. (b) Find the force of compression in the wire.

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Solution

Given:
Radius of the circular wire loop = a
Electric current flowing through the loop = i
Perpendicular magnetic field = B
(a) The force exerted by the magnetic field on a small element dl of the wire,
Fdl = i. (dl × B) = idlB
Using Fleming's left-hand rule, we can say that the direction of magnetic force is towards the centre for any small element of length dl of the wire (dl and B are perpendicular to each other).

(b) Suppose some part of wire loop subtends a small angle 2θ at the centre of a circular loop


So,
2Tsinθ = i.(dl × B)
= i.2aθ B (Using length of any arc, l =rθ)
∵ θ is very small, sinθ = θ
2Tθ = i.2aθB
∴ T = iaB

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