Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

212 of 276

Question 212 of 276

Two 10cm10cm long, straight wires, each carrying a current of 5A5A are kept parallel to each other. If each wire experienced a force of 10−5N{10}^{-5}N, then separation between the wires is _____ cm.cm.

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Answer: 5

Given: F=10−5NF={10}^{-5}N

Force between wires
Fl=μ0i1i22πd\dfrac{F}{l}=\dfrac{{\mu }_{0}{i}_{1}{i}_{2}}{2\pi d}, where, dd is separation between them.
⇒10−50.1=2×10−7×25d\Rightarrow \dfrac{{10}^{-5}}{0.1}=\dfrac{2\times {10}^{-7}\times 25}{d}
⇒d=50×10−3m=5cm\Rightarrow d=50\times {10}^{-3}m=5cm

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