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

Answer:

The current is  $I_b = 400 \ A$

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

        $I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }$

Here  $\mu_o$ is the permeability of free space with value  $\mu_o = 4 \pi * 10^{-7} \ N/A^2$

=>      $I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }$

=>      $I_b = 400 \ A$