Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.7
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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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:
F = μ0 Ia Ib L/2 πr
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
F = (4π x 10⁻⁷ x 8 x 5 x 0.10 )/2 π x 0.04 = 2 x 10⁻⁵
The magnitude of force is 2 x 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.