Question

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current.

Answer 1

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Step-by-step explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

`F = (\mu_oI_1I_2L)/(2\pi r)\\\\I_(2) = (2F\pi r)/(\mu_oI_1L)`

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

`I_(2) = (2F\pi r)/(\mu_oI_1L)\\\\I_(2) = (2(2.4*10^(-4))(\pi)(0.4))/((4\pi*10^(-7))(12.5)(3))\\\\I_(2) = 12.8 \ A`

Therefore, the bottom current is 12.8 A to the right.