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Homework Help: Find EMF - Loop inside a loop generating a magnetic field

  1. Jul 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A loop of wire with an area 1cm^2 is centered within a larger loop of area 1m^2. Apower supply is generating current through the larger loop described by the function:

    I(t) = 40A * cos(2t/s * 2pi/5)

    Calculate EMF at time t = 3s

    2. Relevant equations
    Magnetic flux = BAcos(theta)
    EMF = d/dt Magnetic Flux = d/dt BAcos(theta)
    B = ui/2r = (u/2r) 40A * cos 2t/s * 2pi/5)
    u = permeability of free space = 4pi * 10^-7

    r = sqrt(area of circle/pi)

    3. The attempt at a solution

    Some of the work done above, I set the equation of B into the emf then took the derivative with respect to time.

    EMF = 57 d/dt uA/2r cos(3t - 2pi/5)^2
    EMF = 57*3* uA/2r cos(6t- 2pi/5)

    I confused myself over which A I was referring to and which radius as well. Thinking about this it makes sense that A is the smaller loop's area and r is the bigger loop's radius.

    Does this look correct?
    Last edited: Jul 30, 2014
  2. jcsd
  3. Jul 31, 2014 #2
    No comment? I'm also now questioning if my theta in the magnetic flux is the same theta in the function given...
  4. Jul 31, 2014 #3


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    Hmm... well ##emf = \frac{d\phi}{dt}##, and ##\phi = \vec{B}\cdot \vec{A}## which means you need the magnetic field function within the smaller loop. There's a function for that that involves current and constants. You almost have the function. You're missing a pi in the denominator (I'm pretty sure anyway).

    This makes me think you're misunderstanding the physical interpretation of this function.
    Say I have a current carrying wire, and I want to calculate the magnetic field at SOME POINT, I just plug the current through the loop in, find the distance, plug that in for r, and presto, I have my B field.

    You want the flux, which doesn't consist of 1 point. What you calculated was the flux through dA, where this particular dA is a distance r from a wire. You need to sum up an infinitely many amount of these dA's.

    What you have is pretty close, but what's your 'r' in your equation?

    And to answer your second question, no it's not. The "theta" given (I think you mean t?) doesn't really represent an angle (unless you wanna get into a 4d coordinate system), whereas the cos(theta) from your dot product is the measureable angle between the magnetic field vector and the normal vector from your area.
  5. Jul 31, 2014 #4


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    One more comment about the thetas, since the magnetic field is ultimately an extenction of the lorenz force (which is a cross product) any point that lies in the plane of the area will have a magnetic field that is normal to the area, which means that it is parallel to the normal vector of the area, which means cos(theta)=1
    (it seems as though you got that yourself, as it just kinda dropped out of your equation, but I wanted to be sure)
  6. Jul 31, 2014 #5


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    Also, upon thinking about it, I think there's a few assumptions you have to make in order to solve this problem. One of which is the geometry of the area, another is the orientation of the areas. I think it's safe to assume that they lie in the same plane (as I assumed in post #4). I'm not so sure about the geometry, is this all the information that was given?
  7. Jul 31, 2014 #6
    Oh alright so I(3) = what I want for I in the equation of the magnetic field and from what you're saying I think I have the right B equation. This was my resource (shows how pi drops out).

    http://faculty.wwu.edu/vawter/physicsnet/topics/MagneticField/MFLoops.html [Broken]

    Would r in B = ui/2r be the distance from the large hoop to the small hoop? In other words, radius of large hoop minus radius of small hoop?

    There was also a picture....
    Last edited by a moderator: May 6, 2017
  8. Jul 31, 2014 #7


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    Ok, so they're circles, that makes it a little easier. As for the pi, I thought you were canceling the pi in the denominator with the pi in mu naught.

    Also, read post #3 again. I don't think you're going to get out of integrating your function to get the flux (as for r).

    What is I(3) compared to I(t)? Ultimately, you want dphi/dt or ##\frac{dB \cdot A}{dt}## If you use I(3) in that equation, what will you get? If you use I(t) what will you get? (I'm not asking for an answer, just describe what you'll get with words)

    Notice, that the flux is still a function of r. You need to not have that. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html

    Look at the second section, and pay close attention to the title there. That's the equation that you're using, but it explains where it comes from and what you get out of it. Maybe that will shed some light on this situation, as I'm not really sure how I can go much further without telling you step by step how to solve the problem.

    I might go away for awhile, I have a funny feeling we're about to get really busy at work. I'll check back when I get a chance.
  9. Jul 31, 2014 #8
    If I use i(3) then I'd receive the amount of current at the time t=3. If I use just the equation, I'd probably end up having to take the derivative with respect to time. Is that correct?

    I'm not really aware of any magnetic field equations that would fit this problem without an R term... but I think I understand your point. That equation I was using was B due to a current element in the center. I'm looking for B from the large loop onto the smaller loop.
  10. Jul 31, 2014 #9


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    Exactly. You don't want I(3), you want I'(3) (as I is the only function in your equation that is dependent on time).

    Also, you need the biot-svart law to the best of my knowledge. When you use that equation, the mu naught I/2R, you're getting the magnetic field at the CENTRAL POINT of the loop, you don't want just that. You want the sum of the magnetic fields through all points contained within the smaller loop. In order to do that, what do you need?
  11. Aug 1, 2014 #10
    In order to do that, I'll need an integral containing the entire area of the loop. This just got way more complicated than I thought it would be. How am I supposed to apply this integration in this problem?
  12. Aug 1, 2014 #11


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    Try here.

    By definition ##\phi = \int \int F \cdot n dS## where F is the vector field, here the B field, n is the normal dS is the differential surface area.

    So first off, you need a function for the field w.r.t. distance from the current loop. Once you have that, then you need to integrate that function to sum up all the "field lines" running through the small loop.
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