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Electromagnetic induction graph

  1. Jan 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A small bar magnet is being slowly inserted with constant velocity inside a solenoid as shown in figure. Which graph best represents the relationship between emf induced with time
    1422355278232Capture.PNG

    1422355326547Capture.PNG


    2. Relevant equations
    ε = -dφ/dt
    φ = B.A

    3. The attempt at a solution
    Since the magnet is brought closer to the solenoid at a constant speed, the emf should be constant, then it should fall to zero inside the solenoid and finally an emf in the opposite direction should be generated.
    The closest answer to my reasoning is (C), which is indeed the correct answer.
    However, I don't understand why (C) is sort of a sine wave. I thought it would be more of a constant function.
     
  2. jcsd
  3. Jan 15, 2016 #2

    cnh1995

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    What you guessed is a square wave. The emf is not induced in all turns at the same time. As the magnet proceeds, the flux reaches further turns. So, the emfs induced in the turns are phase shifted. If you add the emfs vectorially(phasor addition), you'll get a sinewave.
     
  4. Jan 15, 2016 #3
    I'm sorry but I've never done something like this. Could you explain some more?
    You're from India, this is an IITJEE question btw.
     
  5. Jan 15, 2016 #4

    cnh1995

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    Consider the first turn. When flux links with the first turn at t=0, emf E1 is induced in the first turn. The next turns have no flux linkage(very less, practically). As the magnet moves ahead, the flux starts linking with the next turns. So, when flux links with 2nd turn, it is already linked with 1rst turn. So, now the emf will be E1+E2, at t=t1. As time passes, the magnet moves ahead and emf is induced in the next turns. So, at t=t3, emf will be E1+E2+E3 and so on. If you plot this graph, you'll get an approximate sine wave.
     
  6. Jan 15, 2016 #5
    When you're entering the first loop, lets say current in the first loop starts flowing anticlockwise (emf: E1). As you exit the loop and move to the second loop, the current in the SECOND loop is anticlockwise (emf: E2) however, since we've exited the first loop, the current in the first loop is now clockwise (emf: - E1). So, at time t = t1, the emf is -E1 + E2?
    What am I doing wrong?
     
  7. Jan 16, 2016 #6

    haruspex

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    You need to consider more loops. As the magnet approaces the first loop, there is current due to all loops, principally from the nearer loops, then gradually less contribution from more distant loops. So during that phase, the current is increasing. When the magnet is half way between the first two loops, they will cancel each other, so the current is now only from the remaining loops, and will be a little less than at the peak. The current reduces steadily until it is half way through the coil, then starts to reverse.
     
  8. Jan 16, 2016 #7

    cnh1995

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    I was visualizing this situation with respect to a similar problem I did a week ago, which involved motional emf in generator stator coils. This problem is different. This is what I think...Consider only two turns. As the magnet moves towards the first turn, the emf increases as the flux increases. When the magnet is inside turn 1, the induced emf will be at it's peak. As magnet starts leaving turn 1, there will be -ve emf -E1 induced in turn 1 and +ve emf E2 induced in turn 2. But E2-E1≠0. When magnet is exactly in the middle of turn 1 and turn 2, E2-E1=0 (i.e. rate of decrease of flux in turn 1=rate of increase of flux in turn 2). After that, E2-E1 will be negative as reduction of flux in turn 1 will be more than increase of flux in turn 2. Hence, the graph of emf will go more and more negative, till the magnet is inside the 2nd turn. This will be the -ve peak point of emf (this is also the instant when the magnet "starts" leaving the 2nd loop). As the magnet leaves the 2nd turn, emf will go on to be less negative and finally, when all the flux is removed, emf will again be 0. I'm not sure if this is correct.
     
    Last edited: Jan 16, 2016
  9. Jan 17, 2016 #8
    I was thinking something similar. Thanks!
     
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