Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the

Question

4 years ago

7

current in the other is 3.45 A.
Find the magnitude of the force per unit length that one wire exerts on the other.

1 answer:

Natasha_Volkova [10] · Answer 1 · 2022-01-25T22:14:05+03:00

The force per unit length between the two wires is $6.0\cdot 10^{-5} N/m$

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0 = 4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

$I_1, I_2$ are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

$I_1 = 1.75 A$

$I_2 = 3.45 A$

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m$

Learn more about current and magnetic fields:

#LearnwithBrainly