[Physics] Current induced in loop moving out of magnetic field : contradiction using Fleming’s right hand rule

electric-currentelectromagnetismhomework-and-exercisesinductionmagnetic fields

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?

Best Answer

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.