A wire loop into a magnetic field

[ian2012]

Imagine pushing a conducting loop (rectangular and rigid and perfectly conducting) into a magnetic field B. The loop is inserted from x<0 into the region x>0 permeated by a uniform magnetic field B=B(z-direction). The loop lies in the x-y plane. The loop is sufficiently long in the x-extent that at all times part of the loop remains in the region x<0.

Now imagine the loop has a resistor R at x<0 and a capacitor C just below the resistor (also at x<0). Imagine that the loop is being pushed along by a constant velocity v. The emf in the loop = emf across the resistor + emf across the capacitor.

It can be shown that the current in the loop will decay with time:

$$I=\frac{BvL}{R}e^{\frac{-t}{RC}}$$

Thus the current decays with time despite the fact that you continue to push it into the magnetic field region. Why does this happen?

I am guessing as the induced current decays as it charges up the capacitor There is a potential difference created across the capacitor. And so since the opposing force depends on the induced current, the opposing force is smaller. Eventually there is no opposing force, the loop will accelerate towards the region x>0.

 

[Bob S]

If the single-turn coil has dimensions x and y, the voltage developed around the loop as the coil is inserted is

V = - d(B·x·y)/dt = -B·y·dx/dt = -B·y·v_x volts

where v_x is a constant velocity in the x direction. Initially, there is a retarding force due to Lenz's law, but this force, which is proportional to the current in the wire, decays away with the RC time constant as the capacitor charges, as you point out. So then there is no force on the loop until the pushing of the coil stops. [STRIKE]The reaction of the coil + capacitor is to continue pulling the coil into the field at the initial velocity until the current, produced by the battery V and the resistor R, dies out with the same RC time constant, and the coil stops. [/STRIKE]This is nearly the same as when a big sheet of pure aluminum or copper is pushed into a magnet and the retarding eddy currents are generated, but in this case there is no capacitor, but there is a small L/R time constant (due to the inductance L of the eddy current loops).

Bob S

[revised] During the initial pushing, the capacitor charges to the voltage V, with a stored energy E_c = ½CV². During the initial push, there is an equal amount of energy dissipated in the resistor. When the current dies out, no further pushing is required to maintain the constant velocity v_x. So after the initial current dies out, there is no Lorentz force, and no opposing pushing force. The loop will maintain this velocity until there is another applied pushing force. The capacitor will remain charged to the voltage V, which is exactly the Faraday-Law-induced voltage from the motion of the loop in the magnetic field, and no current in the loop.