Question

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?

Answer 1

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 .