Download Test 2 Solution - James Madison University

Test 2
University Physics II (E&M, Waves) PHYS250
James Madison University
Spring 2012
NAME_________________________
Honor Pledge: I have not knowingly given or received unauthorized help on this test.
Signature________________________
You may use a calculator or blank spread sheet. NO BOOKS OR NOTES ARE ALLOWED.
Useful formulas and constants:
c = 3.0 x 108 m/s
R = 8.31 J/mol.K
h=6.63 x 10-34 J-s
Pa=1.01x105 Pa
e = 1.6 x 10-19 C
mp=1.007276 u
mn= 1.008665 u
mHY=1.007825 u
J= 4.184 joules/cal
Mega = 106
Cice= 0.5 cal/gm-oC
Cwater= 1 cal/gm-oC
g = 9.8 m/s2
1 cal = 4.186 J
1 eV = 1.6x10-19 J
ρwater = 1 gm/cm3
εo = 8.85 x 10-12 F/m
1 u= 1.66 x 10-27 kg
me= 0.000549 u
1 Ci =3.7x1010 Bq
vo=331 m/s= 1087 ft/s
micro(µ)= 10-6
Calum= 0.22 Cal/gm-C
Csteam= 0.48 cal/gm-oC
NA = 6.02 x 1023 molecules/mole
k = 1.38 x 10-23 J/K (Boltzman)
G=6.67x10-11 m3/kg s2
µo = 1.26 x 10-6 N/A2
v (sound)=340 m/s room temp.
R=1.1097 x 107 m-1 (Rydberg)
k = 9.0 x 109 N-m2/C2 (Coulomb)
1(u) c2 = 931.5 MeV
nano = 10-9
pico = 10-12
LF= 80 cal/gm(fusion, water0)
Lv= 540cal/gm (vap., water)
electron
proton
neutron
qe= -1.6 x 10-19 Coulombs
qp=1.6 x 10-19 Coulombs
qn=0 Coulombs
me = 9.11 x 10-31 kg
mp = 1.67 x 10-27 kg
mn = 1.67 x 10-27 kg
Part 1 (30 points)
Concepts
8
4
10
Part 2 (70 points)
Problems
8
10
15
15
15
15
15
all correct extra
Put your name on all sheets of paper. Use as much paper as you need. If you are including extra work
write - “see other sheet” - so that I find all your work when grading.
CONCEPTS: Be sure to choose the most correct answer
1 State clearly Lenz’s law.
current is generated in a loop to oppose the change in the flux. Flux is the amount
of magnetic field passing through the loop. So the current generates a B field so
that there is no change in the total amount of B. Flux uses Area and B that are
parallel.
2 Name two types of electromotive forces that can result based on Faraday’s law.
Describe how they differ.
Motionalè due to the motion of charge through a field. Based on magnetic force
on a moving charge but can be set equal to a changing flux
Inducedè Follows Faraday’s laws with a changing B field an no magnetic force on
the charges. In this case an electric field is set up.
3 Who discovered the first magnetic monopole?
ANSWER:
Not yet discovered
4 Plot below i(t), VR(t), VL(t), VC(t) for an RLC circuit. You may assume arbitrary
amplitudes. Draw and lable your plot carefully. Be sure that the plots line up
vertically.
Also gaave credit if the phasors were shown with the correct orientation. Plotting could mean drawing
a vector although this seems to me to be a stretch I still decided to give full credit for the correct
phasors.
Kevin Giovanetti - PHYS250_1_2_SP12
Problems from the Whole Course
8
1.20 ohm
18.0 V
17.5 cm
10.8 mT
The right edge of the circuit in the diagram extends into a 10.8 mT uniform magnetic field. What is the magnitude
of the net force (in N) on the circuit?
Tries 0/99
What is the direction of the force?
A. Into the screen
B. Out of the screen
C. Up the screen
D. Down the screen
E. To the left
F. To the right
The direction is F to the right
Tries 0/99
1
The battery provides a voltage V. The resistors are shown in the diagram. What is the current through
the battery. [If you can’t do this problem you may remove one resistor. Draw the new diagram.
Solve this new problem for ½ of the points. Simply remove the resistor. Do not replace it with
a wire.]
Three resitors in corner reduce to a resistor of value R (
series give 2R,
then parallel è (1/2R +1/2R)=1/R
The voltage across the middle is zero. So the effective resistance is 2R along each leg which are in
parallel. So the current is V/R.
2 A very long straight wire carries a current of 1 A moving from left to right. The pictures below show
the wire from the side and looking directly down the wire. Draw the electric and magnetic fields in
both of these pictures. Be careful with your drawing. Do not try to draw something that is vague so
that it may be partially correct.
Electric Field
NO Field
Electric Field
NO Field
Magnetic Field
Magnetic Field
current out of the page
field is circles that are separated by larger
distances as you get away from the wire (field
decreases) Field runs CCW.
O (out)
X (in)
If this is a standard wire in a typical electric circuit (long copper wire with a battery providing the
current) then what is the total charge on the wire?
Total charge on the wire
zero
0
Kevin Giovanetti - PHYS250_1_2_SP12
Homework6
12
3
18.68 cm
4.95 m/s
A potential difference of 0.0405 V is developed across a 18.68 cm long wire as it moves through a magnetic field at
4.95 m/s. The magnetic field is perpendicular to the axis of the wire. What is the strength (in T) of the magnetic
field?
Tries 0/99
What is the direction of the magnetic field?
A. Into the screen
B. Out of the screen
C. Up the screen
D. Down the screen
E. To the left
F. To the right
Tries 0/99
F=qvXB F on +charge to the left, v down, B out of the page.
The voltage or work per unit charge W/q is force times distance for each charge. V=Fl/q =vBl
.0405/(4.95 .1868)
B=00. 0438T out
Kevin Giovanetti - PHYS250_1_2_SP12
Homework7
25
47.1 ohm
(10.0 V)cos(wt)
18.7 mH
16.5 uF
For the circuit shown above, what is the resonance frequency (in rad/s)?
Tries 0/99
For the circuit shown above, what is the resonance frequency (in Hz)?
Tries 0/99
For the circuit shown above, what is the VR (in V) at resonance?
Tries 0/99
For the circuit shown above, what is the VL (in V) at resonance?
Tries 0/99
4
ω=1800 rad/s
f=287 Hz
Vr=47.1Io
Vc=33.66 Io
VL=33.66 Io
capacitor and inductor cancel. V(t)=i(t)R
VR=10
Io=0.212 A
VL=Io ωL=7.15
5
Kevin Giovanetti - PHYS250_1_2_SP12
Homework6
21
The diagram above shows a U-shaped conducting rail that is oriented vertically in a horizontal magnetic field of 0.603
T. The rail has no electrical resistance and does not move. a slide wire with mass, 10.7 g, and resistance, 0.372 Ω, can
slide up and down without friction while maintaining electrical contact with the rail. The slide wire of length, 24.4
cm, is released from rest. What is the terminal speed (in m/s) of the wire? Assume that the local acceleration due to
gravity is 9.80 m/s2.
Tries 0/99
The rail stops increasing speed when Fgravity=Fmagnetic
mg=IlB
I=V/R
V=change in Flux=B vΔt l /Δt= multiply B times the area change l * length down vΔt then
divide by the time.
V=Bvl
I=Bvl/R
mg=(Bvl/R)lB
v=mgR/(B2l2)=1.8m/s