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Question
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
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Solution
Current flowing in wire A, IA = 8.0 A
Current flowing in wire B, IB = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l = 10 cm = 0.1 m
Force exerted on length l due to the magnetic field is given as:
B = `(mu_0 2"I"_"A""I"_"B""I")/(4pi"r")`
Where,
`mu_0` = Permeability of free space = 4π × 10–7 T m A–1
B = `(4pi xx 10^-7 xx 2 xx 8 xx 5 xx 0.1)/(4pi xx 0.04)`
= 2 × 10–5 N
The magnitude of the force is 2 × 10–5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.
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