1. The problem statement A current I passes through a straight conducting wire located within this page. Rectangular circuit ABCD on the page is moving away from the straight conducting wire with speed v, while side AB remains parallel with the current I. The length of side AB is 2a, and the length of side AD is 2b. The resistance of the circuit is R. How much current flows through circuit ABCD when the center O of the circuit is distance r from the straight wire? The magnetic permeability of a vacuum is m, and the magnetic flux density created by the circuit's current is negligible. Please help me!
Hi N_Spears and welcome to PF. Please follow the rules of this forum and use the template when you seek help with homework. Show us the relevant equations and tell us what you tried and what you think about the problem. We just don't give answers away.
Hix, this question isn't my homework, it's included in MEXT scholarship test. I have checked all the formula that I've known but none of them can solve this question. I suspect that our curriculum doesn't contain the right one. I live in Viet Nam. Can you help me?:(
I can help you, but we still don't give answers away. There is no single formula that you can find that will give the answer simply by plugging in. You need to assemble the answer, but first you need to understand what is going on. Answer the following question first. 1. Why should there be a current in the rectangular loop and where does this current come from?
judging from what I've learned, current cannot appear in the rectangular but magnetic flux. And if it does, it will come from the straight conducting wire for sure, this happen because of the current flows into this wire. I was really confused about this question, I think there must be a mistake because as I said before, current cannot appear in the rectangular but magnetic flux.
Is this always true? What if the magnetic flux through the rectangular loop is changing with time? What does Faraday's Law say?
Sorry, you're right. I've just revised my lesson. So, the current appears in the rectangular loop because the magnetic field is changing as the the rectangular loop is moving away from the wire. Please give me the next question, I promise that I will check my book before giving the answer :)
You need to calculate the magnetic flux through the rectangular loop. Note that the magnetic field is not uniform over the surface of the loop.
We have [tex]\Phi = BS\cos\alpha[/tex], but this formula just can be used when the magnetic field is uniform :|, I wonder if there is another formula :-<
Then you can use this one: [tex]d\Phi = \vec{B}\vec{dS}[/tex] I thought this formula was quite popular in most Vietnamese physics textbooks ???
That's the formula. Make yourself a drawing of the rectangle and the wire. Draw a line on the rectangle such that the B field is constant along it. What does it look like? If the thickness of the line is dy (very small) what is its area? What is the element of magnetic flux dΦ through it?
According to my textbook, your formula and mine are the same :D @kuruman: uhm, so it looks like electromagnetic induction of the current flows into the wire affect 2 points in the loop whose distance will be r - b and r + b. And if the thickness of the line is very small so its area will be 2b. Thus, we have [tex] B = \frac{mI}{2\pi(r - b)} + \frac{mI}{2\pi(r + b)} = \frac{mIr}{\pi (r^2 - b^2)} \Longrightarrow d\Phi = \frac{mIr2b}{\pi(r^2 - b^2)} ; (\cos\alpha = 1) [/tex]. Is this right? But I don't know that we can sum two B of 2 points, can we? :|
You didn't pick up my hint. If the length of the line is 2b and its width is dy, what is its area? dA = ???
In most Vietnamese textbooks particularly, S denotes area. But instead, people usually use A internationally (A is the 1st letter of "area", you see).
Correct. The element of area is 2b dy. Call it "dA" or "dS" or "Fred" or whatever you like. The name, as you know, is not important. Now that you have the area element, can you calculate the magnetic flux through that area element? That's what dΦ is. dΦ = ?? This is where you are headed after this step: 1. You need to add all such elements dΦ continuously (i.e. do an integral) to get the total flux through the rectangular loop. 2. You need to take the time derivative dΦ/dt and use the result in Faraday's Law to find the induced emf. 3. You need to use Ohm's Law with the induced emf and the resistance to find the current.