JEE Main 2022 · 28 Jun · Shift 2 · Physics Q43

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PhysicsSingle correct+4 / -1Question 43 of 90

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of $x$ A in the same direction. If the force of attraction per meter of each wire is $2\times10^{-6}$ N, then the value of $x$ is approximately :

2.4

1.4

Answer and worked solution

Official answer: C

Force per length. $\dfrac{\mu_0x^2}{2\pi d}=\dfrac{2\times10^{-7}x^2}{0.2}=10^{-6}x^2=2\times10^{-6}$ , so $x=\sqrt2\approx1.4$ .

(C)

Topics:Force on a current-carrying conductor

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JEE Main 2022 · 28 Jun · Shift 2, Physics question 43