The force of attraction between two long parallel conductors $ 1 \mathrm{~m} $ apart carrying unit ampere current in same direction is

Question

The force of attraction between two long parallel conductors $ 1 \mathrm{~m} $ apart carrying unit ampere current in same direction is, $ 2 \times 10^{-7} \mathrm{Nm}^{-1} $, $ 10^{-7} \mathrm{~N} \mathrm{~m}^{-1} $, $ 10^{-5} \mathrm{~N} \mathrm{~m}^{-1} $, $ 0.5 \times 10^{-6} \mathrm{~N} \mathrm{~m}^{-1} $

MARKS App · Accepted Answer

Given: the distance between parallel conductors d = 1 m , current in both parallel conductor is i 1 = i 2 = 1 A . The force per unit length of attraction between two parallel current-carrying conductors is given by, f = μ 0 i 1 i 2 2 π d f = 2 × 10 - 7 × 1 × 1 1 = 2 × 10 - 7 N m - 1 .

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