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Q.

Two straight parallel coductors varying same current i are arranged horizontally one above the other separated by a distance of 1 mm. The upper wire is hold by a support. The lower wire of linear density 1 gm/m is hanging in air due to magnetic force of current. The value of i is

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a

10 A

b

49 A

c

21A

d

7 A

answer is D.

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Detailed Solution

$\frac{m}{l} = \frac{1 g m}{m} = 10^{- 3} k g / m$
$i_{1} = i_{2} = i, r = 1 m m = 10^{- 3} m$
$\begin{array}{l} F = m g \\ \frac{μ i_{1} i_{2} l}{2 π r} = m g \\ \frac{{}_{0}^{μ} i^{2}}{2 π} = \frac{m}{l} g r \\ \frac{4 π \times 10^{- 7} \times i^{2}}{2 π} = 10^{- 3} \times 9.8 \times 10^{- 3} \\ i^{2} = \frac{9.8 \times 10^{- 6}}{2 \times 10^{- 7}} = 49 \\ i = \sqrt{49} = 7 A \end{array}$

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