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Can current be induced in a uniform magnetic field?

  1. Apr 22, 2015 #1
    Currently doing my A2 Levels and have just been revising over electromagnetic induction. The general consensus seems to be that current can only be induced in a changing magnetic flux.The text book I'm using constantly refers to cutting a magnetic field, i.e. moving a wire relative to a field, to induce current. But the only way I can see this working is if the field is non uniform. If a wire were to cut a non uniform field (by moving in a straight line), since magnetic field strength is different along its course of motion, then the magnetic flux 'felt' by the wire is changing, thus meeting to requirements for electromagnetic induction. However, if a wire was to cut a uniform magnetic field (that is going from left to right say), by moving its length downwards through the magnetic field, surely no current would be induced, since the field strength remains constant along its course of motion and so magnetic flux does not change. In fact, my text book actually has questions that refer to this example: e.g. 'The uniform field between the poles of a magnet has flux density of 0.062T. A wire of length 5.3cm is moved down a distance of 2.8cm in 0.060s. Calculate the EMF and the current induced'. I understand how to work through this question, using faraday law etc, but I don't understand why any EMF/current would be induced in the first place for the reasons explained previously.
    So its either the idea that current can only be induced via changing magnetic flux is wrong, because it can also be induced by just moving a wire relative to a uniform field, OR the questions in my text book are just poorly formulated.
    I think its the latter but I'm not sure. Any help?
  2. jcsd
  3. Apr 22, 2015 #2
    yeah, I too find this electromagnetic stuff tricky.

    For this example, I wonder if it would help to think of it not as a single isolated wire in a uniform field, but as a piece of wire length that is part of a loop...a loop that must be there to close the electric circuit...then, as the wire moves through a uniform field, the amount of flux linked by the loop does change...and it is this change that induces a current.
  4. Apr 23, 2015 #3
    I don't really understand, why does flux change if its in a loop?
  5. Apr 23, 2015 #4


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    Staff: Mentor

    In this situation, the concept of flux doesn't apply, because the wire doesn't form a closed loop. In order to talk about flux, you have to have a closed loop.

    The concept that applies here is motional emf: ##\mathcal{E} = Blv## if the wire and its velocity are perpendicular to B.


    In the diagram on that page, the stationary U-shaped wire and the moving rod create a loop so that a current can flow. If you remove the U-shaped wire, leaving only the moving rod, you still have an emf between the ends of the rod. It doesn't produce a current because you don't have a closed circuit.

    With the stationary U-shaped wire in place, you now have a complete loop. You can calculate the total emf around the loop in two ways:

    1. The U-shaped wire is stationary, so its three sides have zero motional emf. The total emf around the loop equals the motional emf produced in the moving rod, which we can calculate as above.

    2. We can calculate the rate at which the area of the loop changes, and use Faraday's law: $$\mathcal{E} = d\Phi_B/dt = d(BA)/dt = B(dA/dt) = B(d(lx)/dt) = Bl(dx/dt) = Blv$$ (the same result as in method 1).

    Suppose instead of this setup, you have a rigid loop with fixed area A, and move this whole loop through the field with constant velocity v to the right. Now, both the left and right sides have an upwards motional emf. With respect to the loop, the emf is counterclockwise on the left side and clockwise on the right, so the net emf around the loop is zero. And Faraday's law also gives zero because both B and A are constant.

    You might now wonder, is Faraday's law just a shortcut method for evaluating the net motional emf around a loop? No! Consider a stationary loop, but now let the magnitude of B increase or decrease at a rate dB/dt. There is no motion, so there is no motional emf. But there is nevertheless an emf around the loop which you can calculate using Faraday's law! $$\mathcal{E} = d(AB)/dt = (dB/dt)A$$
    Last edited: Apr 23, 2015
  6. May 1, 2015 #5
    Simple. When a conductor is being moved in UMF then current is induced. Faraday's law. Emf=Blv
  7. May 1, 2015 #6
    The Wikipedia article on Faraday's Law has a very interesting section on when the law does not apply.

    I mean, personally, it's one of those things I'm not 100% certain about either. How does the circuiut wire know about the total amount of magnetic field it is encompassing? Because in the end, it's electrons who must feel the force which then create the current.
    For example, what if you increased the magnetic field selectively in the center, while keeping it the same where it intersects with the wire? How could the circuit possibly know about that change?
    The only solution I can think of is that you can't modify a field in that way, since "squeezing" more field lines through the center would cause the ones at the wire to be pushed out. That would be like moving the wire, which creates a Lorentz force on the electrons again.

    EDIT: Thinking more about this, clearly there are also limits on the frequency of the change. If I send a photon through the center of the circuit, Faraday's Law would say that at some point there should be an induction happening since a photon is after all a magnetic wave. But, it's rather intuitively obvious that shining a flashlight through the middle of my circuit won't do anything. So clearly the law is an approximation.
    Last edited: May 1, 2015
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