Two thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a current Io=200A going into the page. A third thin, infinitely long wire with mass per unit length λ=5g/m carries current I going out of the page. What is the value of the current I in amps in this third wire if it is levitated above the first two wires at height h=10cm above them and at a horizontal position midway between them?

Question

Two thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a current Io=200A going into the page. A third thin, infinitely long wire with mass per unit length λ=5g/m carries current I going out of the page. What is the value of the current I in amps in this third wire if it is levitated above the first two wires at height h=10cm above them and at a horizontal position midway between them?

Answer 1

Two thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a currentTwo thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a currentTwo thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a currentTwo thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a currentTwo thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a currentTwo thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a current

Answer 2

To determine the value of the current I in the third wire, we can use the principle of magnetic forces between currents. The magnetic force between two parallel wires is given by the equation:

F = (μ₀ * I₁ * I₂ * L) / (2 * π * d)

where F is the force between the wires, μ₀ is the permeability of free space, I₁ and I₂ are the currents in the first and second wires, L is the length of the wires, and d is the distance between them.

In this case, the force acting on the third wire is balanced by the gravitational force, because it is levitated above the first two wires. The magnetic force on the third wire is given by:

F = (μ₀ * I * Io * L) / (2 * π * d)

where I is the current in the third wire.

Since the force is given by F = mg, where m is the mass per unit length, and g is the acceleration due to gravity, we can equate the two forces:

(μ₀ * I * Io * L) / (2 * π * d) = m * g

Rearranging for the current I:

I = (2 * π * d * m * g) / (μ₀ * Io * L)

Substituting the given values, we have:

I = (2 * π * 0.03 * 0.05 * 9.8) / (4π × 10^-7 * 200 * 1)

Simplifying the equation, we get:

I = 29.4 A

Therefore, the value of the current I in the third wire is 29.4 Amps.

Anyone?

To determine the value of the current I in the third wire, we can use the principle of magnetic forces between currents. The magnetic force between two parallel wires is given by the equation: