. A current of 21A is passed

Question

through a 20m length of two
conducting wire placed 125cm
apart. Determine the force of
attraction between the wires
assuming the current is flowing in
the direction.
A. 1.33 x 10-5N
B. 4.5 x 10-5N
C. 7.1 x 10-5N
D. 6.4 x 10-5N

Answer 1

To determine the force of attraction between the wires, we can use the formula:

Force = (μ₀ * I₁ * I₂ * L) / (2 * π * d)

Where:
- Force is the force of attraction between the wires,
- μ₀ is the permeability of free space, equal to 4π x 10^(-7) N/A²,
- I₁ is the current passing through the first wire,
- I₂ is the current passing through the second wire,
- L is the length of the wire,
- d is the distance between the wires.

In this case, the current passing through both wires is the same, so we can substitute I₁ = I₂ = 21A. The length of the wire is given as 20m, and the distance between the wires is given as 125cm = 1.25m.

Plugging these values into the formula:

Force = (4π x 10^(-7) N/A² * 21A * 21A * 20m) / (2 * π * 1.25m)
= (4 * 21² * 10^(-7) N * m / A²) / (2 * 1.25)

Simplifying the expression:

Force = (4 * 441 * 10^(-7)) / 2.5
= 1764 * 10^(-7) / 2.5
= 7056 * 10^(-8) / 25
= 282.24 * 10^(-8) N
= 2.8224 * 10^(-5) N

Rounding the result to two decimal places, we get:

Force ≈ 2.82 x 10^(-5) N

The correct answer is not listed in the given options.