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 Author Topic: RC circuit + magnetic induction (B field)  (Read 9175 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: December 12, 2009, 01:26:11 pm » posted from:Taipei,T\'ai-pei,Taiwan

A capacitor has been charged to $V_o$ , a metal bar with mass m ,resistor R is located between two parallel wire (distance L) in magnetic field B, as shown in the above figure.
At t=0; the switch was turn from a to b.
Current I will flow throught metal bar, which F=I L B will accelerate meta bar. $m\frac{dv}{dt}=I L B$.
When the metal bar is moving, the changing in magnetic flux will induce voltage $V_i= B L v$
So the equation for the loop is $Vc= \frac{Q_c}{C} = I R + B L v$, where $I=-\frac{dQ_c}{dt}$

$\frac{1}{C} \frac{dQ_c}{dt}=-I =R \frac{dI}{dt}+B L \frac{dv}{dt}= R \frac{dI}{dt}+B L \frac{BLI}{m}$
, $\frac{dI}{dt}=-(\frac{1}{RC}+\frac{B^2L^2}{mR}) I$
the solution is $I(t)=\frac{V_o}{R}(1-e^{-\alpha t})$ ,where $\alpha=\frac{1}{RC}+\frac{B^2L^2}{mR}$

The following is a simulation for the above case:
The charge $Q_c(t)$, the velocity $v(t)$ and the current $I(t)$ are shown (C=1 in the calculation).

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Never underestimate others. Never overestimate oneself. ...Wisdom