Download Physics 142 Lecture 19

Quiz# 2
deals with answering questions about
(and calculating values for)
charged particle beams being accelerated and deflected by
E and B fields.
A recap of some in-class problems/questions dealing with
charged particles accelerating across a potential difference
and/or being steered or deflected by Electric or Magnetic fields
is posted for you at
http://csce.unl.edu/~ddaes/Quiz2Review.ppt
You might also want to review the textbook’s
Assigned problems:
Ch.19: P 56, 76
Ch.20: P 17,18
Ch.22: P 17
If there is an AC current in a wire loop,
there is a changing B field within the loop.
Does this changing B field
induce an EMF (voltage) of
its own in the loop of wire?
1) Yes, it must
2) No, it does not.
B field
1) Yes, it must
AC current produces a changing B field.
A changing B field induces an EMF in a loop, even the loop that
produces the changing B field
This is called self-inductance
Self-Induction in Daily Life
When you turn off your toaster by unplugging it
 current changes rapidly to zero
 rapidly changing B field produced
 very large voltage induced in wires of toaster
 spark
voltage difference large enough
for electrons to flow through air
If you unplug your computer while it is
still on, large voltages may be generated,
possibly destroying the CPU chip.
A solenoid has an inductance of 87.5mH and
a resistance of 0.250. Find:
The time constant for this circuit:
 = L/R = 87.510-3H/0.250 = 0.350 seconds
and how long it takes for the current to reach ½
its maximum once connected to a 3volt battery.
6A = 12A(1-e-t/)
½ = e-t/
ln(½) = -t/0.350sec
t = 0.2426 sec
How much energy is stored in this solenoid at full current?
A lamp with a 60W bulb is plugged into a 120v outlet.
a)What are the rms and peak currents through the lamp?
b)What is the bulb’s resistance?
A lamp with a 60W bulb is plugged into a 120v outlet.
a)What are the rms and peak currents through the lamp?
P  I rms R
2
but requires a value for R
b)What is the bulb’s resistance?
A lamp with a 60W bulb is plugged into a 120v outlet.
a)What are the rms and peak currents through the lamp?
P  I rms R but requires a value for R
P  I rmsVrms
Irms=60W/120v = 0.5 A
2
b)What is the bulb’s resistance?
A lamp with a 60W bulb is plugged into a 120v outlet.
a)What are the rms and peak currents through the lamp?
P  I rms R but requires a value for R
P  I rmsVrms
Irms=60W/120v = 0.5 A
1
while I rms 
I peak so I peak  2I rms  1.414  0.5A
2
2
b)What is the bulb’s resistance?