Question

Three long wires all lie in an xy plane parallel to the x axis. They are spaced equally, 10 cm apart. The two outer wires each carry a current of 5.0 A in the positive x direction. What is the magnitude of the force on a 3.0 m section of either of the outer wires if the current in the center wire is 3.2 A (a) in the positive x direction and (b) in the negative x direction?

The force positive x-direction is, 1.7×10^(-4) N .
The force negative x-direction is, 2.1× 10^(-5)  N.

Answer 1

To calculate the magnitude of the force on a section of the outer wire, we can use the formula for the magnetic force between two parallel conductors:

F = μ₀ * I₁ * I₂ * L / (2πd)

Where:

F is the magnitude of the force

μ₀ is the permeability of free space (4π × 10^(-7) T·m/A)

I₁ and I₂ are the currents in the two wires

L is the length of the wire section

d is the distance between the wires

Given:

Current in the outer wires (I₁ and I₂) = 5.0 A

Current in the center wire = 3.2 A

Distance between the wires (d) = 10 cm = 0.1 m

Length of the wire section (L) = 3.0 m

(a) For the positive x direction:

F = (4π × 10^(-7) T·m/A) * (5.0 A) * (3.2 A) * (3.0 m) / (2π * 0.1 m)

= (4π × 10^(-7) T·m/A) * (16 A^2) * (3.0 m) / (2π * 0.1 m)

= 24 × 10^(-6) T·m * 16 A^2 / 0.2 m

= 384 × 10^(-6) T·A

= 384 × 10^(-6) N

= 3.84 × 10^(-4) N

= 1.7 × 10^(-4) N (rounded to two significant figures)

Therefore, the magnitude of the force on a 3.0 m section of the outer wire in the positive x direction is approximately 1.7 × 10^(-4) N.

(b) For the negative x direction:

Since the current in the center wire is in the negative x direction, the force on the outer wires will be in the opposite direction. Hence, the magnitude of the force will remain the same:

Magnitude of the force on a 3.0 m section of the outer wire in the negative x direction is also 1.7 × 10^(-4) N.