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Faraday's law theoretical question

  1. Sep 10, 2015 #1
    This is just a general theoretical question.

    If i have a piece of wood that is wrapped with coils and then run a current through these coils. Will the magnetic field generated by this cause a voltage in a conductor if it is close enough to it?

    If this is the case could you explain how that would work, if not could you provide an explanation.

    Many thanks
     
    Last edited by a moderator: Sep 10, 2015
  2. jcsd
  3. Sep 10, 2015 #2

    Zondrina

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    So you have some tightly wound conducting coils. You run a current through these coils, which produces magnetic flux ##\phi_{B_1}## through the central region of the inductor.

    If we place a second coil near the first coil, a current ##i_1## in the first coil produces a magnetic flux ##\phi_{B_2}## through the second coil. If we change ##\phi_{B_2}## by changing the current, an induced emf appears in the second coil.

    Note I'm talking about mutual induction when I say the aforementioned. If we change the current with time in the first coil, an induced emf appears in the second coil due to the changing flux. We can define the mutual inductance of coil 2 with respect to coil 1 as:

    $$M_{21} = \frac{N_2 \phi_{B_{21}}}{i_1} = \frac{N_2}{i_1} \iint_S \vec B_1 \cdot d \vec S_2$$

    Where ##N \phi_B## is the magnetic flux linkage. We can then define the induced emf across coil 2:

    $$\varepsilon_{21} = - M_{21} \frac{d i_1}{dt} = - \left[ \frac{N_2}{i_1} \iint_{S_2} \vec B_1 \cdot d \vec S_2 \right] \frac{d i_1}{dt} $$

    A similar thought process will allow one to deduce the induced emf in the first coil:

    $$\varepsilon_{12} = - M_{12} \frac{d i_2}{dt} = - \left[ \frac{N_1}{i_2} \iint_{S_1} \vec B_2 \cdot d \vec S_1 \right] \frac{d i_2}{dt} $$
     
  4. Sep 10, 2015 #3
    what if I run an AC current through it instead of a DC current. the first coils that is?
    will that ensure a constant changing magnetic field thus inducing voltage in second coil?
     
  5. Sep 10, 2015 #4

    Zondrina

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    Yes, the changing current in the first coil causes a changing magnetic flux in the second coil, which induces an emf in the second coil to drive an induced current.

    The induced current in the second coil will vary with time because of the changing magnetic flux through the second coil. Since the induced current varies as well, it in turn will cause a changing magnetic flux through the first coil. The thought process continues as you would expect.
     
  6. Sep 10, 2015 #5
    ok so what if the second coil was undergoing corrosion - say a decrease in cross section thickness.
    will the voltage then drop in the second coil and then could i measure this?

    i'm thinking with faraday's law?
     
  7. Sep 10, 2015 #6

    Zondrina

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    Remember, the induced emf ##\varepsilon_{L_2}## in the second coil can be found using Lenz's law. The emf always opposes the induced current.

    Suppose a current ##i_2(t)## is induced in the second coil by the first coil. Lenz's law gives the direction of the induced emf, and the emf itself can be calculated using:

    $$\varepsilon_{L_2} = - L_2 \frac{di_1}{dt} = - \frac{N_2 \phi_{B_{21}}}{i_1} \frac{di_1}{dt} $$

    Where the inductance ##L_2## is actually a mutual inductance ##M_{21}##.

    We use Faraday's law to calculate the self-induced emf, not the mutually-induced emf.
     
    Last edited: Sep 10, 2015
  8. Sep 10, 2015 #7
    now just to clarify (sorry electrical theory is not my strong point) when you say emf you're talking about induced voltage?

    also could i relate the induced emf in the second coil to a formula including the area of this coil?
     
  9. Sep 10, 2015 #8

    Zondrina

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    The term emf is actually depreciated now, but it does refer to a voltage.

    One more thing though. If we happen to know the magnetic flux through the second coil caused by the changing current in the first coil, then we could hypothetically use Faraday's law to calculate the induced emf instead (even though we are talking about mutual-inductance rather than self-inductance).
     
  10. Sep 10, 2015 #9
    so if i ran an AC current through the first coil there is absolutely no justification for using faraday's law on the second coil? unless we knew the flux in the second coil? and that is impossible to know?
     
    Last edited: Sep 10, 2015
  11. Sep 10, 2015 #10
    i just updated my post. last question you had already answered apologies for that.
     
  12. Sep 10, 2015 #11

    Zondrina

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    Okay so lets put two coils next to each other.

    Next, run an AC current through the first coil so the current in the first coil will constantly be changing. Note: current is the movement of electric charges in time.

    A magnetic field will be produced by the moving electric charges. By looking at each individual electron spin comprising the current, you will be able to see why a magnetic field is produced.

    The magnetic field that is produced in this case will vary in strength over time because the AC current producing it is also varying. Physically, the state of the electron spins is changing over time.

    Since the strength of the magnetic field is changing in the first coil, a changing magnetic flux is produced in the second coil. Due to the changing magnetic flux in the second coil, an induced emf is produced that drives an induced current. You may calculate the induced emf ##\varepsilon_{21}## by using the equation in my prior post.
     
  13. Sep 10, 2015 #12
    could you configure the ac current so that flux could essentially be considered constant? like say get a flux value that just bounces from one value to another repetitively? for instance 1 and then -1 then 1 again and so on, but moving in a discrete manner?
    Thanks for the help so far appreciate it.
     
  14. Sep 10, 2015 #13

    Zondrina

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    You could use a ##120 V \space \text{rms}, 60 Hz## AC line to feed an AC signal through a power transformer. This signal can then be fed through a diode rectifier, filter, and voltage regulator to massage the signal however you want. The output signal can then be taken across a load.

    In essence, this would be the process used to convert AC to DC. With a DC current, the strength of the magnetic field in the first coil would be constant and not vary with time. So there would be no emf produced in the second coil as a result because the magnetic flux through the second coil would not be changing.

    Clearly, you don't want to use a DC current here.

    If you use AC current, the flux value will be bouncing from one value to another anyway. I would recommend a square wave AC signal for your purposes to make the flux change in a discrete manner.
     
  15. Sep 10, 2015 #14

    Zondrina

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    To summarize the information for you:

    - Suppose you have a single coil in a closed loop circuit with an AC power source. Now turn the AC source on, which will cause AC current to flow through the circuit containing the coil. The current will vary in strength, so the magnetic field produced in the central region of the coil will vary in intensity. This change in the magnetic field intensity will cause a change in the magnetic flux passing through the coil. This changing magnetic flux will cause a self-induced emf that will drive a self-induced current. The self-induced emf can be found using Faraday's law.

    - Suppose you have two coils together now with the first coil in a closed loop circuit with an AC power source. Turn the AC source on in the first circuit, which will cause a changing magnetic field intensity as before. The changing magnetic field in the first circuit causes a changing magnetic flux in the second coil. This changing magnetic flux leads to a mutually-induced emf and mutually-induced current. The mutually-induced emf can be found using ##\varepsilon_{21}##.
     
  16. Sep 10, 2015 #15
    thank you Zondrina.
     
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