# 2 parallel rods carrying the same current

## Homework Statement

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is 0.70 m, while the mass of each is 0.071 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 8.2 10-3 m. Determine the current in the rods.

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## Homework Equations

F = ILBsin(theta)
B = (uI)/(2 pi r)

## The Attempt at a Solution

i know that current will be the same for both rods and since the current is flowing in the same direction, the magnetic force will be attractive between the rods. i know the length of each wire and the mass. i dont know how to progress in this problem because it seems to me that i need to know the magnetic field to be able to solve either equation. is there a way to solve this problem without knowing the magnetic field or am i overlooking something here? I'm stumped.

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ranger
Gold Member
Well you can have two equations with two unknowns, and it is possible to find both unknowns. Its kind of like solving for x and y with two algebraic equations. Only difference is that here, we have current and the magnetic field.

i'm sorry but is there possibly another way to say what you just told me; i don't get it. Am i even using the correct formulas? if i try to use F = ILB sin(theta) and solve for I, i get stuck because i don't know how to find B. If i try to solve B=(uI)/(2 pi r) i run into the same situation.. i dont know I or B. I know this problem cannot be as hard as it seems to me.....

ranger
Gold Member
Okay so since the rods are not falling towards the earth or each other, that must mean that the force between the rods is equal to that of gravity. So we have:
mg = ILB sin(theta)
B = (uI)/(2 pi r)

We know all the values for all variables there, except B and I. Now using a system of equations, can you solve for both B and I? You can use the method of substitution or elimination, whatever makes you happy.

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ok, i'll give it another shot. thank you.