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Faraday's Law and the radius of the coil

  1. Sep 5, 2014 #1
    I've only recently started learning basic electrodynamics, but I don't understand why a loop of coil with a small area and a magnet falling through will produce a larger peak to peak emf amplitude than a loop of coil with a larger area with the same magnetic field falling through.

    To clarify, lets say you have a loop of coil and you drop a magnet through the coil. This is will produce an emf according to Faraday's law, which is a function of time since the magnet is falling, but let's say you measure the peak to peak amplitude of this function. Now if you have a loop of coil with a larger radius, and you drop the same magnet through at the same speed, it will produce a smaller value for the peak to peak emf. Why is this?
     
  2. jcsd
  3. Sep 5, 2014 #2

    Philip Wood

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    Gold Member

    I'd written you an explanation but it's disappeared (clicked something wrongly I suppose). Haven't time to write it out again, but the thumbnails appended are the main thing. It's all about change in flux linked with the coil as the magnet drops. The drawings show the magnet close to the coil, where the flux linked should be near its maximum, but in the lower diagram, only lines L1 and R1 are linked. This means that the line and the coil are intertwined like adjacent links in a chain. The other lines "go back on themselves" within the cross-section of the coil. In the top diagram, L1,L2, R1, R2 are all linked.

    More formally, we evaluate the net flux through the cross-section of the coil as [itex]\int \vec{B}.\text{d} \vec{A}[/itex]. When the magnet is near the coil and we'd want a large flux linkage, for the large coil the flux linkage will be low because the flux is going in one direction in the inner zones of the coil's cross-sectionl, and in the opposite direction in the outer zones, so the integral doesn't amount to much.
     

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    Last edited: Sep 5, 2014
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