- #1
ian2012
- 80
- 0
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:
[tex]I=\frac{BvL}{R}e^{\frac{-t}{RC}}[/tex]
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.
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:
[tex]I=\frac{BvL}{R}e^{\frac{-t}{RC}}[/tex]
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.