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lbvjy [14]
3 years ago
7

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

current in the other is 3.45 A.
Find the magnitude of the force per unit length that one wire exerts on the other.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

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:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

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