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3 years ago

15

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of

the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?

1 answer:

lions [1.4K]3 years ago

6 0

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i² = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

= 0.2 A approx .

2 i = .4 A Ans .

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