W9_parallel_resonance.eps

264 Week 8: Faraday’s Law and Induction

or
dvx
dt

+

B^2 L^2 vx

mR

= 0 (576)

which is the usual first order, linear, homogeneous ordinary differential equation and is trivially
integrable (see remarks in the math review section if the following doesn’t make perfect sense to
you):

dvx

dt =

(

−

B^2 L^2

mR

)

vx

dvx

vx

=

(

−

B^2 L^2

mR

)

dt

ln(vx) =

∫ dv

x

vx

=

∫(

−

B^2 L^2

mR

)

dt

ln(vx) =

(

−

B^2 L^2

mR

)

t+C

vx(t) = v 0 exp(

(

−

B^2 L^2 t

mR

)

(577)

(^00) 0.5 1 1.5 2 2.5 3
0.2
0.4
0.6
0.8
1
t
v(t)
Figure 93: A plot of the exponential decay of the velocity of the rodas its initial kinetic energy
is “burned” in heating the resistor with the induction-driven current that also slows it down. The
units of the plot arev 0 (forv) andτ=BmR (^2) L 2 (fort).
A plot ofvx(t) is shown in figure 93 in units ofv 0 and the exponential decay timeτ=BmR (^2) L 2.
The magnetically induced electrical voltage produces a current that produces a force in the
magnetic field thatslows the rod down. If energy is indeed conserved, we would expect that the rate
at which the kinetic energy of the rod decreases shouldexactly matchthe rate at which Joule heating
from the current occurs in the resistor. That way the negative work done by the induction force is
precisely balanced by the positive appearance of heat energy in theresistor throughout; energy isn’t
being created, it is just being changed from one form to another.
This is easy enough to test algebraically. The rate at which power appears in the resistor is
(substututing in several results from above):

PR=I^2 R=

B^2 L^2 v^2

R =

B^2 L^2 v^20

R exp(

(

−

2 B^2 L^2 t

mR

)

(578)