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Q.

Two long wires have uniform charge density $λ$ per unit length each. The wires are non-coplanar and mutually perpendicular. Shortest distance between them is d. The Interaction force between them is:

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a

$\frac{2 λ^{2}}{ε_{0}}$

b

$\frac{λ^{2}}{2 π ε_{0}}$

c

$\frac{λ^{2}}{4 ε_{0}}$

d

$\frac{λ^{2}}{2 ε_{0}}$

answer is A.

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Detailed Solution

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$F = \int d F = \int E \cos θ d q = \int E \cos θ (λ . d y) = \int \frac{λ}{2 π \in_{0} r} \cos θ λ . d y N o w, \frac{y}{d} = \tan θ ⇒ d y = d . \sec^{2} θ d θ A l s o, \frac{d}{r} = \cos θ \Rightarrow r = d . \sec θ ∴ F = \int d F = \int \frac{λ}{2 π \in_{0} d \sec θ} \cos θ λ d \sec^{2} θ d θ \Rightarrow F = \frac{λ^{2}}{2 π \in_{0}} \int_{- \frac{π}{2}}^{\frac{π}{2}} d θ = \frac{λ^{2}}{2 \in_{0}}$

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Two long wires have uniform charge density λ per unit length each. The wires are non-coplanar and mutually perpendicular. Shortest distance between them is d. The Interaction force between them is: