Recent content by j.m.g.

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

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?

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.

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.

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...

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...

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

The ball drops due to the force of gravity.

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

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.

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...