Question

Two straight long parallel wires, separated by a distance of 1.0 m, carry a current of 4.00
each. A third wire is arranged perpendicular to the two parallel wires. If the magnetic field at a point equidistant from all the wires is zero, find the current in the third wire.

Answer 1

The magnitude of the current in the third wire is 8 A.

Total magnetic field

E₁ + E₂ + E₃ = 0

Equidistance from the two wires = 0.5 m

Apply Biot Savart rule to determine the magnitude of the magnetic field.

μI/2πr₁ + μI/2πr₂ + μI/2πr₃ = 0

(4π x 10⁻⁷ x 4)/(2π x 0.5) + (4π x 10⁻⁷ x 4)/(2π x 0.5) + (4π x 10⁻⁷ x I₃)/(2π x 0.5) = 0

4 x 10⁻⁷ I₃ = -3.2 x 10⁻⁶

I₃  = (-3.2 x 10⁻⁶)/(4 x 10⁻⁷)

I₃ = - 8 A

Thus, the magnitude of the current in the third wire is 8 A.

Learn more about magnetic field here: https://brainly.com/question/7802337

#SPJ1