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NCERT Class XII Chapter
Moving Charges and Magnetism
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Question : 7 of 28
Marks: +1, -0
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.
Solution:  
Force of attraction per unit length on two parallel wires carrying current in same direction.
F/l = μ04π2I1I2r\frac{\mu_0}{4\pi} \frac{2I_1I_2}{r} = 10−710^{-7} × 2×8×54×10−2\frac{2\times8\times5}{4\times10^{-2}} = 20 × 10−510^{-5} N m−1m^{-1}
Attractive force on 10 cm section of wire A
F = [20 × 10−510^{-5}] [10 × 10−210^{-2}] = 2 × 10−510^{-5} N
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