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Is lenz law violated?

  1. Aug 4, 2008 #1
    1. suppose a long solenoid is connected to a DC source and after a long time steady state has reached. And suddenly an iron rod is inserted in the solenoid along its axis. Obviously, self inductance of the solenoid will increase by some factor. [One can assume some resistance of of the solenoid if required].
    MY question is:
    (a) what happens to the magnetic field at the center of the solenoid.
    (b) what happens to the magnetic flux through an infinitesimally small area at the center of solenoid.

    My answer to them are:
    1(a) magnetic field reduces - as current will reduce.
    1(b) flux doesn't change. - as lenz's law says that induced emf is such that it opposes the cause. so flux shouldn't change.

    2. suppose a long solenoid is connected to a constant current source and after a long time steady state has reached. And suddenly an iron rod is inserted in the solenoid along its axis. Obviously, self inductance of the solenoid will increase by some factor. [One can assume some resistance of of the solenoid if required].
    MY question is:
    (a) what happens to the magnetic field at the center of the solenoid.
    (b) what happens to the magnetic flux through an infinitesimally small area at the center of solenoid.
    My answer to them are:
    2(a) magnetic field reduces - as current will reduce. [but how can current reduce if we are using a constant current source across an inductor (i am confused here)].
    2(b) flux doesn't change. - as lenz's law says that induced emf is such that it opposes the cause. so flux shouldn't change.

    Can some one comment of both the answers and correction please if i am saying something wrong.
     
  2. jcsd
  3. Aug 4, 2008 #2
    H won't change much, but B will increase greatly. Since iron has a relative permeability much greater than air, and B = mu*H, and H is determined by the amp-turns divided by the integral magnetic path length, B will increase very much. Any e/m fields reference text will affirm this. I hope I've helped. BR.

    Claude
     
  4. Aug 5, 2008 #3
    1.
    i am little confused as I thought that if the solenoid is connected to a battery; finally when the steady state is reached, then the current through the solenoid will be i = V/R.
    Now, when we insert the iron rod, flux through the solenoid will not change immediately, while the inductance will increase by some factor 'n'.
    And flux = L i. L: inductance i: steady current

    Thus, L i = nL*i' Therefore, i' = i/n
    Thus, just after inserting the solenoid, the current reduces, and so the magnetic field due to the solenoid reduces as well.
    Once again, after a long time the current through the solenoid will again become V/R.

    Now, due to the current the iron rod gets magnetized permanently. And the magnetic field due to iron rod is much higher than the magnetic field due to the solenoid.
    Thus, the total magnetic field increases to a high value.
    My final point is:
    a) magnetic field due to the current in the solenoid will reduce initially and come back to the same value finally.
    b) magnetic field due to the iron rod will remain permanently, and thus the overall affect will be that the magnetic field will increase.
    Am i going wrong somewhere in my concepts? Please help.
     
    Last edited: Aug 5, 2008
  5. Aug 5, 2008 #4
    But you keep saying "magnetic field"? Are you referring to "B", the magnetic flux density, "phi" the total flux (phi = A*B)), or "H", the magnetic field intensity. The steady state current is related directly to H. The I and H before and after are equal after transients have been dissipated.

    When you insert the iron rod, you are doing work, as you will feel a force. Your force integrated over the distance equals the work done. The B will increase due to the high mu of the iron rod. Thus the total energy before is B*H/2, or mu0*(H^2)/2. The total energy after is mu0*mur*(H^2)/2, where mu0 is the free space permeability and mur is the relative permeability of the iron rod.

    The energy after is much greater than before due to the work done inserting the rod. Lenz' law is not violated at all. When the rod is moving through the H/B field, any induced mmf/emf is oriented oppositely. The momentary decrease in current is due to the increase in volt seconds. Since the power source was specified as a constant voltage type (battery), any increase in the inductance voltage drop results in a decrease in voltage across the winding resistance, hence a momentary decrease in current.

    I believe these types of problems are solved as examples in classic physics texts such as Halliday-Resnick and other similar texts. Does this help? BR.
     
    Last edited: Aug 5, 2008
  6. Aug 5, 2008 #5
    yes surely.. this helped me a lot. ... and yes i was actually reading halliday resnick and so i was little confused abt this point..
    moreover i was not using proper notations so my question went a bit confusing...
    i was looking for the net magnetic field just after inserting the iron rod.. and of course it would reduce. But with time it will rise to a very high value.
     
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