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enter image description here

In the picture shown magnetic field exists into the plane (denotes by the X) and the loop (iii) is moving out of the region of the magnetic field.

According to Lenz's law, the induced current will counter the change in magnetic flux that produced it, and hence the induced current should be clockwise to counter the change in magnetic flux. That is what I'm getting when I apply the right hand thumb rule to obtain the direction of the induced current.

Similarly Fleming's right hand rule (which I believe gives the direction of the induced current) should also give the answer to be clockwise. enter image description here But when I apply the rule, my finger points to the bottom right corner of the screen. So according to the path which is available to the charges (governed by the shape of the loop) I find that the current in the lower part i.e. adc of the loop, the current works out to be anticlockwise. But according to the path available to the current, it works out to be clockwise for the upper part of the loop i.e. abc.

Why is there are contradiction? Am I using the rule incorrectly?

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    $\begingroup$ Think that you need to make your reasoning and question clearer. $\endgroup$
    – user93237
    Commented Mar 12, 2017 at 21:41
  • $\begingroup$ I agree with Sam Weir : your reasoning is vague. Please show show how the induced current "turns out to be different directions for the upper and lower parts of the loop." $\endgroup$
    – sammy gerbil
    Commented Mar 12, 2017 at 23:12
  • $\begingroup$ @sammygerbil I've edited my question. I apologise for the inconvenience caused by the improper phrasing of my question. $\endgroup$
    – Kunal Pawar
    Commented Mar 13, 2017 at 3:09
  • $\begingroup$ You say that the loop (iii) is moving "out of the plane". But the arrow attached to that loop shows the loop moving down and to the left, which is in the plane. Is that right? If not, and the loop really is moving out of the plane (opposite to the direction of the magnetic field), then there's no change in flux and thus no current. $\endgroup$
    – Mike
    Commented Mar 13, 2017 at 3:30
  • $\begingroup$ @Mike I think Kunal means that the loop is moving out of the region of the magnetic field. $\endgroup$
    – sammy gerbil
    Commented Mar 13, 2017 at 15:39

1 Answer 1

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When the loop is wholly within the region of the uniform magnetic field, then there is no induced current even when the loop is moving. As you pointed out, in this case the induced currents on opposite sides of the loop are opposite and equal so they cancel out.

It is only when one side of the loop leaves the region of the magnetic field, or more generally enters a region in which the field is different, that the induced currents are no longer equal. In this case there is no induced current in the part of the loop which is no longer within the magnetic field. More generally, the side of the loop on which the magnetic field is stronger will determine the direction of the resultant current.

If the loop is moving out of the region of magnetic field with a uniform speed, would the induced current be steady? If not, why? Because of the loop's geometry?

Good follow-up question.

Induced emf (and therefore also induced current) is equal to rate of change of flux linkage through the loop. If there is a sudden change in flux linkage, the current will change suddenly.
enter image description here
If the loop is a rectangle with one side parallel to the edge of the magnetic field, then there is a sudden change in flux linkage as that edge crosses the boundary, and again when the trailing edge crosses. In between there is a constant decrease in flux linkage, therefore a constant current. The current-time graph is rectangular.

For both a circular loop and a rectangular loop with a corner crossing the boundary first, the current will increase and decrease continuously from zero to a maximum, because the flux linkage is not changing suddenly. For the circle the induced current does not change uniformly, because of the non-linear edges; the current-time graph is semi-eliptical. For the rectangular corner=first loop the current-time graph is triangular or trapezoidal. For both circular and rectangular corner-first loops, the the maximum current occurs when the widest part of the loop crosses the boundary, because that is when the flux changes most rapidly.

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  • $\begingroup$ Well if the loop is moving out of the region of magnetic field with a uniform speed, would the induced current be steady? If not, why? Because of the loop's geometry? $\endgroup$
    – Kunal Pawar
    Commented Mar 13, 2017 at 16:48
  • $\begingroup$ One more thing. The geometry of the loop will not be fixed (as we move it out of the region of the magnetic field) if the loop is made of something we can easily bend right? Because of Lorentz force? $\endgroup$
    – Kunal Pawar
    Commented Mar 14, 2017 at 1:00
  • $\begingroup$ Yes, if the wire loop is not rigid the magnetic forces on it could bend it. $\endgroup$
    – sammy gerbil
    Commented Mar 14, 2017 at 1:56
  • $\begingroup$ Just remembered you said you were gonna expand your answer... still waiting. $\endgroup$
    – Kunal Pawar
    Commented Mar 25, 2017 at 20:26

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