A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is :

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$1 .2 \times 10^{- 4}$ N directed away from wire X.

$1.2 \times 10^{- 5}$ N directed towards wire X.

$1.2 \times 10^{- 4} N$ directed towards wire X.

$2.4 \times 10^{- 5} N$ directed towards wire X.

answer is A.

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Detailed Solution

Force of interaction $= I_{1} ℓ_{1} B_{12}$
$\begin{array}{l} = \frac{μ_{0} I_{1} I_{2}}{2 πr} ℓ_{1} \\ = \frac{4 π \times 10^{- 7} \times 6 \times 0 .5}{2 π \times 5 \times 10^{- 2}} \\ = 1 .2 \times 10^{- 5} towards X \end{array}$

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