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Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F₁. If distance between wires is halved and currents on them are doubled, force F₂ on 10 cm length of wire P will be :

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8F₁

F₁/8

F₁/10

10F₁

answer is B.

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Detailed Solution

$B_{1} = \frac{μ_{0} i}{2 πd} F_{1} = B_{1} i_{2} ℓ = \frac{μ_{0} i^{2} ℓ}{2 πd}$

$∴ F_{2} = \frac{μ_{0} (2 i)^{2} ℓ}{2 π (d / 2)} = 8 \frac{μ_{0} i^{2} ℓ}{2 πd}$
$\Rightarrow F_{2} = 8 F_{1}$

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