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Title: RC circuit + magnetic induction (B field)Post by: Fu-Kwun Hwang on December 12, 2009, 01:26:11 pm
(http://forum.phy.ntnu.edu.tw/neditor/popups/pics/20091211_1593304203.png)
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). |