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Q.

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 $2 \times 10^{- 6} N$ , then the value of x is approximately :

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a

2.4

b

2

c

1

d

1.4

answer is C.

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Force per unit length $= \frac{μ_{0} i_{1} i_{2}}{2 πd}$

$\begin{array}{l} = \frac{μ_{0} \cdot x^{2}}{2 π \times 0 .2} \\ F = 2 \times 10^{- 6} = \frac{4 π \times 10^{- 7} \times x^{2}}{2 π \times 0 .2} \\ \Rightarrow 10^{- 6} = 10^{- 7} \frac{x^{2}}{0 .2} \\ \Rightarrow x^{2} = 10 \times 0 .2 = 2 \\ \Rightarrow x = \sqrt{2} \approx 1 .4 A \end{array}$

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