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Author Topic: RC circuit + magnetic induction (B field)  (Read 8106 times)
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Fu-Kwun Hwang
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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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