Simplified reasoning: current induced in ring so as to oppose change,
dipole field of opposite polarity gives rise to repulsion.
The current will be induced so that the field which would be created by the current loop will oppose the change in flux through it as the loop is removed from the field.
If B is uniform and perpendicular to the flat rectangular loop, the magnetic flux is simply BA = Blx. Thus the induced emf is Blv since the time rate of change of the magnetic flux will be proportional to the time rate of change of x; this derivative is the speed of withdrawal.
If the resistance of the loop is R, then the induced current is Blv/R.
Are we getting something for nothing? Is the current free? No...
Considering the directions of the forces on the segments of the conducting loop, we find that the vertical forces balance. Only F2 is unbalanced, directed to the right with magnitude ilB = (BlV/R)lB = B2l2v/R.
To withdraw the loop from the magnetic field at a constant velocity (no acceleration) will require an external agent to apply a force equal to F2, but opposite in direction. The power supplied by the outside agent is P = F.v = (B2l2v/R) v = (Blv)2/R.
Now if we look at the power dissipated by the resistance of the loop we find that it is exactly the same as the power provided by the external agent; P = i2R = (Blv/R)2 R = (BLv)2/R!
So removal of the loop converts mechanical energy into electrical energy, which is subsequently converted into thermal energy by dissipation by the resistance. Energy is conserved (once again!).
Please note that the external agent moving the loop always experiences an opposing force, ralated to Lenz's law. To confirm this, reverse the velocity so that the external agent is pushing the loop into the field and repeat the arguments above.
Last updated Nov. 4, 1997.
Copyright George Watson, Univ. of Delaware, 1997.