Two long parallel wires carry currents of 2.61 A and 4.05 A. The magnitude of the force per unit length acting on each wire is $3.53 \times 10^{-5}$ N/m. Find the separation distance $d$ of the wires expressed in millimeters.\[d = \, \text{________ mm}\]

Answered: Two long parallel wires carry currents…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Two long parallel wires carry currents of 2.61 A and 4.05 A. The magnitude of the force per unit length acting on each wire is $3.53 \times 10^{-5}$ N/m. Find the separation distance $d$ of the wires expressed in millimeters.

\[d = \, \text{________ mm}\]

Expert Solution

Step 1: Given and formula used :-

We know that the force between the two wires is due to the current flow in the wire as

$straight F space equals space fraction numerator straight mu subscript 0 straight I subscript 1 straight I subscript 2 over denominator 2 pi d end fraction$

where I₁ is the current in the first wire.
I₂ is the current in the second wire.
$straight mu subscript 0$ is the permeability of free space.
d is the distance between the wire.

We know that we have given that Two long parallel wires carry currents of 2.61 A and 4.05 A. The magnitude of the force per unit length acting on each wire is 3.53×10^-5 N/m.

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