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Q.

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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a

8F1

b

F1/8

c

F1/10

d

10F1

answer is B.

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Detailed Solution

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B1=μ0i2πd F1=B1i2ℓ=μ0i2ℓ2πd 

∴F2=μ0(2i)2ℓ2π(d/2)=8μ0i2ℓ2πd
⇒F2=8 F1

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