1. Two straight long parallel wires, separated by a distance of 1.0 m, carry a current of 4.00
A 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.
To find the current in the third wire, we need to apply Ampere's law.
Ampere's law states that the integral of the magnetic field (B) dot product with a small differential length (dl) around a closed loop is equal to the product of the current (I) passing through the loop and the permeability of free space (μ₀).
Mathematically, it can be written as:
∮ B · dl = μ₀ * I
Since the magnetic field at the point equidistant from all the wires is zero, it means that the magnetic fields produced by the two parallel wires cancel each other out at that point.
Now, let's consider the third wire. The magnetic field at the point equidistant from all the wires will be perpendicular to the plane formed by the two parallel wires. To cancel out the magnetic field produced by the two parallel wires, the third wire must produce a magnetic field in the opposite direction.
We can consider a rectangular loop that encloses the third wire. Since the magnetic field is zero at the point equidistant from all the wires, the integral of B · dl around this loop will also be zero.
Therefore, we can write:
B₁ * l₁ + B₂ * l₂ - B₃ * l₃ - B₃ * l₄ = 0
Here, B₁ and B₂ are the magnitudes of the magnetic fields produced by the two parallel wires, l₁ and l₂ are the lengths enclosing the third wire, and l₃ and l₄ are the lengths enclosing the two parallel wires.
Since the wires are long and parallel, their magnetic fields can be approximated as:
B₁ = B₂ = μ₀ * I₁ / (2 * π * r)
where I₁ is the current in the first wire and r is the distance from the wire.
Now, let's consider the lengths enclosing the wires:
l₁ = l₃ = r
l₂ = l₄ = √(r² + 1²)
Substituting these values into the equation and simplifying, we get:
μ₀ * I₁ / (2 * π * r) * r + μ₀ * I₁ / (2 * π * r) * √(r² + 1²) - B₃ * √(r² + 1²) - B₃ * r = 0
Simplifying further:
I₁ + I₁ * √(r² + 1²) / r - 2 * π * r * B₃ - B₃ * √(r² + 1²) = 0
Now, we can solve this equation for the current in the third wire, I₃. Rearranging the terms, we get:
I₁ * √(r² + 1²) = (2 * π * r - √(r² + 1²)) * B₃
Finally, we can solve for I₃:
I₃ = -I₁ * √(r² + 1²) / (2 * π * r - √(r² + 1²))
So, to find the current in the third wire, you would need to know the values of I₁ - current in the parallel wires, and the distance (r) between the wires.