Currents in Parallel Homework: Determine Magnitude of Current

[j.m.g.]

Homework Statement

A long horizontal wire carries a current of 48 A. A second wire, made of 2.8 mm diameter copper wire and parallel to the first, is kept in suspension magnetically 15 cm. Determine the magnitude of the current in the lower wire.

Homework Equations

I know that F/L=(μ0*I1*I2)/(2∏r).

The Attempt at a Solution

By using that equation I am not sure how to solve for F/L. Otherwise I know μ0=4∏*10^-7, I1=48A and r=0.15 m. I also know that I am solving for I2. My main problem is the left side of the equation.

 

[gneill]

What force on the wire is balancing the one due to the current such that the wire is suspended (and not flying away!).

 

[j.m.g.]

I'm not sure. I literally gave you everything I know from the problem. I do know that somehow you need to use F=ma to get the other side to work and F/L ends up being something like ALp, but I am not sure how that works.

 

[PeterO]

I'm not sure. I literally gave you everything I know from the problem. I do know that somehow you need to use F=ma to get the other side to work and F/L ends up being something like ALp, but I am not sure how that works.

To support gneill's question. What would happen if the 48A current was switched off? and why?

 

[gneill]

Putting aside the given problem for a moment, if you wanted to suspend a wire or anything else for that matter, in the air, what force do you need to oppose?

 

[j.m.g.]

You need an equal and opposite force. When looking at the problem I know that the two wires share this equal and opposite force.

 

[gneill]

Suppose I 'suspend' a ball by holding it with my hand a few centimeters above the ground. I release the ball. What happens and why?

 

[j.m.g.]

The ball drops due to the force of gravity.

 

[gneill]

The ball drops due to the force of gravity.

Bingo! The wire is being pushed up (or held up) by the electromagnetic force due to the currents running through them. But what force is it being held against? How might you calculate it? (You may need to look up some additional information based upon the information given in the problem)

 

[j.m.g.]

I am not sure I know how to figure out that calculation. That has been my main problem with this question.

 

[gneill]

I am not sure I know how to figure out that calculation. That has been my main problem with this question.

What information are you given about the object being held up?

 

[j.m.g.]

I have the question I put in the first box and a diagram that shows the current of the top wire is 48A the distance between the two wires is 0.15 m to the unknown wire. I am to find the magnitude of the lower wire. My professor went over this question in class as did my AI and I just cannot grasp the problem for some reason. I understand pretty much everything but how to work the left side of the equation I posted in the first box.

 

[gneill]

You need to figure out the what force (per unit length) that the electromagnetic force needs to balance. Once you have that you can use your equation to determine the required current.

 

[j.m.g.]

Since we are not given a force it was my understanding that somehow we need to plug in B (the magnetic field) times the current times the length. That equals mass times gravity. I have attempted this problem many times, but for some reason I cannot come up with the correct answer. Is there something I am doing wrong?

 

[gneill]

Since we are not given a force it was my understanding that somehow we need to plug in B (the magnetic field) times the current times the length. That equals mass times gravity. I have attempted this problem many times, but for some reason I cannot come up with the correct answer. Is there something I am doing wrong?

Post your calculation so that we can see your attempt.

 

[j.m.g.]

I used ALP=(μ0*I1*I2)/(2∏r). In my equation I put in (1.25∏*10^-7)(8.96*10^3)(9.8)=((4∏*10^-7)(48)I2)/(2∏*0.15). I then solved for I2.

 

[gneill]

I used ALP=(μ0*I1*I2)/(2∏r). In my equation I put in (1.25∏*10^-7)(8.96*10^3)(9.8)=((4∏*10^-7)(48)I2)/(2∏*0.15). I then solved for I2.

What's the 1.25∏*10^-7 number on the left?

 

[j.m.g.]

That was me attempting to figure out A and L. I do think that is my main problem with this equation. I think the rest of my information is correct.

 

[gneill]

Mass = ρ*Area*Length

So that

Mass/Length = ρ*Area

That is, the mass per unit length is the density multiplied by the cross sectional area.

 

[j.m.g.]

Okay. So I used ∏r^2 (∏(2.8*10^-3)^2) for the cross sectional area. I multiplied it by p (8.96*10^3) and by gravity (9.8). I still did not get the correct answer for I2. What am I doing wrong? Is my cross sectional area of a cylinder incorrect?

 

[gneill]

Isn't 2.8mm the diameter and not the radius?

 

[j.m.g.]

You're right, I can't believe I messed that up! Thanks so much for all of your help!