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Homework Help: Magnitude and Direction of Induced Current on a wire around a cylindrical volume

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A uniform magnetic field B is confined to a cylindrical volume of radius 0.080m. B is directed into the plane of the paper and is increasing at a constant rate of (delta)B/(delta)t=0.300 T/s. Calculate the magnitude and direction of the current induced in a circular wire ring of radius 0.16m and resistance 1.4 Ohms that encircles the magnetic field region.

    2. Relevant equations

    3. The attempt at a solution
    step1: find emf through wire ring using magnetic flux of cylinder?
    step2: R(resistance)=1.4Ohms, use equation I=emf/R to find magnitude. and direction is going to be counter clockwise to appose change in flux.

    Not sure how to go about starting on this problem. Help please.
  2. jcsd
  3. May 2, 2012 #2
    You could start by finding the flux phi:
    [itex] \Phi = \int \vec{B}\cdot \text{d}\vec{A}[/itex]
    But already you know how much the flux changes pr. time so:
    [itex] \varepsilon = -\dfrac{\text{d}\Phi}{\text{d}t} = -\dfrac{\text{d}}{\text{d}t}\int \vec{B}\cdot \text{d}\vec{A} = [/itex]
    ? You know that the area dosen't change with time and you could switch the order of integration and differentiation?
  4. May 2, 2012 #3
    Do i already know how much the flux changes or how much the magnitude of the magnetic field changes?

    I keep confusing myself and I dont even know where to start anymore
  5. May 2, 2012 #4
    You know dB/dt and that the magnetic field is uniform, and that you can change the order of diff and int cause the integral is time-independent. So you know how much B is changing and how much the flux canges because you also know your area(i assume that the B-field is parallel with the z axis in the cylindrer)
    Last edited: May 2, 2012
  6. May 2, 2012 #5
    yes, my Area vector is also parallel and in the same direction of B.
  7. May 2, 2012 #6
    ε=(dB/dt)(∏r^2)cos(0°) is that looing right? and im unsure about what to do with the different radius measurements too.
  8. May 2, 2012 #7
    Okay i can't see why you want to do something about the radius but except for that you divide the whole thing by R and assume quasistationary current so it is okay to use: emf = RI
    So what is the problem, you know all the things, just plug in.
    dB/dt = 0.300T/s, A = Pi x 0.080^2m^2.
    There is only 1 radius, the cylinder's radius
    Also there is a sign error i get as a final result:
    [itex] I = \dfrac{\varepsilon}{R} = -\dfrac{\dfrac{\text{d}B}{\text{d}t} \pi r^2}{R}[/itex]
    You are aware that cos(0deg) = 1 right?
    As a final remark: Use Lentz' Law to find the direction of current.
  9. May 2, 2012 #8
    so the radius of the wire circling the cylinder plays no part in the equation? and yes i am aware haha. the negative current in the end identifies the direction as being counterclockwise, am i correct?
  10. May 2, 2012 #9
    When you use Ampere's Law you make up an imaginary loop and the radius is only relevant for [itex]\int d\vec{A}[/itex]
    It is an approximation though.
    That is a bigger loop gives bigger current.
    Yes i believe you are correct: The nature will always try to oppose a change in magnetic flux hence current runs counterclockwise as indicated on the sign of the EMF.
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