Question

A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is the magnitude of the current in the other wire

Answer 1

The current is
`I_b = 400 \ A`

Step-by-step explanation:

From the question we are told that

The length of the segment is
`l = 2.50 \ m`

The current is
`I_a = 1000 \ A`

The force felt is
`F = 4.0 \ N`

The distance of the second wire is
`d = 5.0 \ cm = 0.05 \ m`

Generally the current on the second wire is mathematically represented as

`I_b = (2 \pi * r * F )/( l * \mu_o * I_a )`

Here
`\mu_o` is the permeability of free space with value
`\mu_o = 4 \pi * 10^(-7) \ N/A^2`

=>
`I_b = (2 * 3.142 * 0.05 * 4 )/( 2.50 * 4\pi *10^(-7) * 1000 )`

=>
`I_b = 400 \ A`