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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.
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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