   Chapter 19, Problem 57P

Chapter
Section
Textbook Problem

A wire with a weight per unit length of 0.080 N/m is suspended directly above a second wire. The top wire carries a current of 30.0 A, and the bottom wire carries a current of 60.0 A. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion.

To determine
The distance of separation between the wires so that the top wire will be held in place by repulsion.

Explanation

Given info: The weight per unit length of the top wire is 0.080Nm-1 . The current in the top wire is 30.0A and the bottom wire is carrying a current of 60.0A .

Explanation:

If the system of the two wires has to be in equilibrium, the repulsive force per unit length on the top wire due to the bottom wire has to be equal to the weight per unit length of the top wire.

Fl=μ0I1I22πd

• Fl is the weight per unit length
• I1 is the current in the top wire
• I2 is the current in the bottom wire
• d is the distance between the wires
• μ0 is the permeability of free space

On re-arrangement,

d=μ0I1I22π(Fl)

Substitute 4π×107TmA-1 for μ0 , 0

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