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Capacitor and a moving bar in magnetic field

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Assume that a metal bar of length h and mass m can slide on a rail without friction. The rail completes the circuit through a capacitor of capacitance C. There ia a uniform magnetic field of magnitude B, perpendicular to the plane of the circuit. Assume the resistance of the circuit is negligible, and initially the bar is moving with velocity v. Find the velocity and position of the bar as a function of time. (You can also solve this problem if the resistance of the circuit is R, it is not more complicated.)

    [On the paper, B is given with a circle with a dot at the center, so it points inwards through the paper.]

    3. The attempt at a solution
    The time derivative of the flux through the circuit is hv, and I know that this is the emf of the moving bar. Also, the force on the bar h.I.B. So, I need to find the current and substitute it in F=ma -> h.B.I = m.(dv/dt) to find v(t), and integrate it to find x(t)... But my problem is that I can't find the current through a capacitor with an applied emf of hv.
  2. jcsd
  3. Mar 28, 2008 #2


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    Actually, the resistance is important to the problem, since we are probably going to need to know the induced current. Also, a dot in a circle means the magnetic field is pointing up out of the page.

    The rate of magnetic flux change is Bhv (since the flux is the integral of field times area). The current will then be Bhv/R . You have to use a resistance somewhere in the problem: you can assign it to the moving bar, if you like...

    Keep that capacitor in mind, because they put it there to complicate the problem. At what point will it block the induced current flow?
    Last edited: Mar 28, 2008
  4. Mar 28, 2008 #3
    Sorry, I was too sleepy :)
    I can see that I = Bhv/R when the capacitor is not charged and fully conducting. When it is charged to q = CBhv, it will block the current. How can I reach dq/dt from this?
  5. Mar 29, 2008 #4
    Thinking about this problem, I feel that it will be an oscillator. Because the increase in magnetic flux will generate a clockwise current in the loop (assuming v is to the right), eventually the capacitor will build up a potential difference that opposes the motion of the bar, at which point the bar will stop and turn around. The subsequent decrease in flux will again generate a current, but this time in the opposite direction, so once the capacitor is sufficiently discharged the induced current will overpower it.

    If you ignore resistance, it will be an oscillator. If not, it will be a damped oscillator. {EDIT}In fact, I don't know that you could meaningfully solve this without some resistance.

    One way to solve is to think about energy. The initial energy is all kinetic 1/2 m v^2, and assuming the bar finally comes to rest and the loop has negligible resistance, the final energy is all stored in the capacitor 1/2 C V^2. (Note that v and V are not the same.) You can find equations in your textbook (in the oscillators section) that relate energy to the amplitude and frequency of motion, and from there you can get the equations of motion.

    Or, you can set up Kirchhoff's loop rule. As mentioned, the induced current is Bhv/R, the potential drop across the resistor is IR, and the charge of the capacitor is Q = CV. Since current I is the time derivative of charge Q, Kirchhoff's rules will give you a first order differential equation for Q, the solution of which is a sine/cosine. That implies the motion will be oscillatory also.

    Magnetic force here is IhB, or (B^2 h^2 v )/ R and this can be plugged into Newton's 2nd Law, which will again be a first-order differential equation for v. Be careful about positive and negative signs.
    Last edited: Mar 29, 2008
  6. Mar 29, 2008 #5


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    [EDIT: On further consideration, I will withdraw this remark. I'm leaving it here only to clarify the replies...]

    There is also a critical value for the velocity. It is possible to push the bar fast enough initially so that the velocity oscillates, but the average is greater than zero; otherwise, the bar will rock back and forth. (An analogy would be a pendulum on a pivot. Below a critical value for the velocity at the bottom of the swing, it will oscillate. Above that value, it will spin completely around the pivot with varying speed.)
    Last edited: Mar 29, 2008
  7. Mar 29, 2008 #6
    Isn't there also a critical value if you take damping into account? If I remember correctly, for certain initial conditions the bar's velocity will oscillate but for others its velocity will exponentially decay to zero. Critical damping it's called, or something like that.
  8. Mar 29, 2008 #7


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    Yes, I think that would be true. On the other hand, I'm going to withdraw my earlier remark. As I thought about some more, I realized we probably can't reach a situation where the capacitor is unable to absorb enough energy from the inductive sytem; so we should always end up with oscillatory motion. (I think the situation with resistance in the circuit is more interesting, but it may be easier to analyze the system first without including it.)
  9. Mar 30, 2008 #8
    Ok, this is what I got:
    Kirchoff's Rule: Bhv - q/C - IR = 0, and Newton's 2nd: (dq/dt)hB = m(dv/dt).
    Substituting v from the second eq. I have:
    Aq = dq/dt,
    where A = (h^2 B^2/mR) - 1/RC.
    Now, the solution q(t) is (q_0)exp(At), which goes to infinity with increasing t... Where is my mistake?
  10. Mar 30, 2008 #9
    Let me rethink this for a minute--

    1) In your solution, the coefficient A would not necessarily be positive, in which case you could get an exponential decay q function which decreases to zero, which would make sense since energy is always being dissipated through the resistor as the bar slides. But the problem doesn't really ask for q(t), so...

    2) Go back to where you plugged into Newton's second law. Substitute the induced current (-Bhv/R) into the force expression IhB, and you get [tex]F = -B^{2}h^{2}v/R[/tex] (the negative sign because the force opposes the bar's motion). So then you have a simple differential equation with the expected exponential decay to zero velocity. As far as the sine/cosine stuff I talked about above--

    3) I need to fix my previous post - in order to obtain the sine/cosine behavior, the self-inductance of the loop would also have to be considered. This would add another term to the Kirchhoff equation, -L(dI/dt) and then what you have is actually a SECOND order differential equation for Q, which is hard to solve unless you've taken a course in DEq. Whether the Q function is an exponential or a sine/cosine depends on the physical constants C, R and L. In order to go any further with this, you would need to know L, which is determined by the geometry of the current loop. So I think you can ignore this.
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