Question

Two parallel wires are separated by 0. 06 m, each carrying 3 a of current in the same direction. What is the magnitude of the force per unit length between the wires?

Answer 1

The magnitude of the force per unit length between the wires is $3 \times 10^{-5} N/m$.

What is the magnitude of the force?

In a wire carrying a current I 2, the force that each wire feels due to the presence of the other depends on the distance between them and the magnitude of the currents.

Force per unit length = magnetic permeability * (current 1) * (current 2) / (2 π distance between the wires).

The below expression for the force per unit length.

Moreover, before using the below formula we have to change the unit centimeter into a meter. So, we just divide the centimeter by 100.

$\rm \dfrac{Force}{length}=\dfrac{u_0 \times i _1 \times i_2}{2\pi d}\\\\\dfrac{Force}{length}=\dfrac{u_0 \times i _1 \times i_2}{2\pi d}\\\\Where; \ \mu_0 = 4\pi \times 10^{-7} , i_13 . \ i_2 = 3 \ d=0.06$

Substitute all the values in the formula

$\rm\dfrac{Force}{length}=\dfrac{u_0 \times i _1 \times i_2}{2\pi d}\\\\Where; \ \mu_0 = 4\pi \times 10^{-7} , i_13 . \ i_2 = 3 \ d=0.06\\\\ \dfrac{Force}{length}=\dfrac{4\pi \times 10^{-7} \times 3 \times3}{2\pi \times 0.06}\\\\ \dfrac{Force}{length}= 3 \times 10^{-5} N/m$

Hence, the magnitude of the force per unit length between the wires is $3 \times 10^{-5} N/m$.

Learn more about force here;

brainly.com/question/17035283

#SPJ4