Magnetic Effects & Magnetism

JEE Main level, single correct, medium

210 of 276

Question 210 of 276

Two parallel, long wires are kept 0.20m0.20m apart in vacuum, each carrying current of xAxA in the same direction. If the force of attraction per meter of each wire is 2×10−6N2\times {10}^{-6}N, then the value of xx is approximately

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Answer: (C) 1.41.4

Force per unit length between two parallel wires is Fl=μ0i1i22πd\dfrac{F}{l}=\dfrac{{\mu }_{0}{i}_{1}{i}_{2}}{2\pi d}
Here, i1=i2=i{i}_{1}={i}_{2}=i and Fl=2×10−6N\dfrac{F}{l}=2\times {10}^{-6}N
Then, 2×10−6=4π2×10−7i22π×0.22\times {10}^{-6}=\dfrac{4{\pi }^{2}\times {10}^{-7}{i}^{2}}{2\pi \times 0.2}
⇒i2=2⇒i=1.4A\Rightarrow {i}^{2}=\sqrt{2}\Rightarrow i=1.4A
Hence, x=1.4x=1.4.

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