A vertical straight wire carrying an upward 12-A current exerts an attractive force per unit length of 8.8 x 10-4 N/m on a second parallel wire 7.0cm away. What current (magnitude and direction) flows in the second wire

Magnitude of current in the second wire = 25.67 A

Direction of the current in the second wire Is upwards. (since the force between the two wires is an attractive force, with current directed in the same direction).

Explanation:

Since the force is an attractive force, the current in the second wire is in the same direction as the current in first wire.

The force of attraction between two current carrying wires caerying curremts of magnitude I₁ and I₂ at some distance r, apart is given as

F = (μ₀ I₁ I₂ L)/(2πr)

(F/L) = (μ₀ I₁ I₂)/(2πr)

(F/L) = Force per unit length = (8.8 × 10⁻⁴) N/m

μ₀ = magnetic constant or permeability constant = (4π × 10⁻⁷) H/m

I₁ = current in wire 1 = 12 A

I₂ = current in wire 2 = ?

r = distance between wires = 7.0 cm = 0.07 m

(8.8 × 10⁻⁴) = (4π × 10⁻⁷ × 12 × I₂)/(2π×0.07)

I₂ = (8.8 × 10⁻⁴ × 2π × 0.07)/(4π × 10⁻⁷ × 12)

I₂ = 25.67 A

Therefore, magnitude of current = 25.67 A

Direction of the current in the second wire Is upwards. (since the force between the two wires is an attractive force, with current directed in the same direction).

Hope this Helps!!!

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