Download Physics 203 Exam 1

Physics 203
Gothard Grey
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Coulomb
Weber
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Amp
Volt
Farad
Henry
Ohm
Tesla
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Exam 1
Spring 2008
Name _________________
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Coulomb
C
Js/C
Weber
Js/Wb
Wb
Amp
As
J/A
Volt
J/V
Vs
Tesla
Js/(Tm2)
Tm2
C/s
J/C
C2/J
Js2/C2
Js/C2
Js/(Cm2)
J/Wb
Wb/s
Js2/Wb2
Wb2/J
Wb2/(Js)
Wb/m2
A
W/A
A2s/W
J/A2
W/A2
N/Am
W/V
V
J/V2
V2s/W
V2/W, V/A
Vs/m2
N/(Tm)
Tm2/s
Js2/(Tm2)2
Tm2/A
Tm2/(As)
T
Mass of proton = 1.67E-27 kg
Mass of alpha = 4 x mass of proton
Charge of alpha = +4e
e = 1.6E-19 C
Mass of electron = 9.11E-31 kg
other
V/m = N/C
s/
s
1E5 Gauss
o = 4E-7 Tm/A
o = 8.85E-12 C2/Nm2
c = 3.0 m/s
A proton, a neutron and an electron enter a magnetic field. The proton shouts to the
others “You guys are going the wrong way!” “Are we behind the curve on this?” the
electron asked. The neutron replied “I’m neutral on this, I won’t get bent out of shape
about it since I’m not in charge anyway.” The proton circled around and said “Look, I’m
positive!”
Superposition principle:
Magnetic fields from multiple sources add (as do waves, electric fields, gravity fields, ....)
Magnetic forces on moving electric charge:
F = qv x B
F = IL x B
p = Ber
Electric forces on charges:
F = qE
Magnetic field from electric current:
∫closed path Bds = μo Ienclosed = μo ∫ j  dA
∫ dA = area enclosed by path
dB = μo/(4πr2) Ids x r ; r = unit vector from source point to
point where B is being calculated.
Used mainly to determine direction and if no simpler method exists.
Magnetic fields worth remembering:
Current source:
Field:
Long straight wire
B = μoI/(2πr)
Coil with N turns
B = μoNI/(2r)
Solenoid (long coil)
B = μoNI/l = μonI n = N/l
Toroid
B = μoNI/(2πr)
Magnetic Flux:
Induced EMF:
dA = Area
 = N d/dt 
direction so that magnetic field from induced current opposes .
Waves: c = 1/(oo)
 =  B dA
Inductance:
L = oN2A/l for solenoid
V across inductor = -L d/dt I, across capacitor, V = - Q/C
Energy stored in an inductor = ½ L I2 where I is the maximum current.
Energy stored in a capacitor = Q2/2C where Q is the maximum charge stored.
Transformers:
V1/N1 = V2/N2
V1I1 = V2I2
Multiple Choice 3 points each
[1] Which of the following will NOT induce a potential in a conducting loop in a
magnetic field?
(a) The strength of the magnetic field is changed.
(b) The loop is rotated about an axis perpendicular to the magnetic field.
(c) A very strong magnet is placed in the center of a conducting loop, at rest.
(d) The size of the loop is changed.
Magnetic flux is not changing in c.
[2] Magnetic field lines
(a) end on magnetic poles.
(b) trail behind moving charges.
(c) are always perpendicular to electric field lines. (d) close without ending.
This is one major difference between magnetic and electric fields ( no
monopoles).
[3] You are given an iron bar, A, and a rather weak magnet, B. For which of the
following arrangements would you detect no force?
(C) The magnet only attracts objects to the poles, not to the center.
[4] A magnetic field exerts a force on all the following EXCEPT
(a) a current carrying wire
(b) a magnet
(c) a moving electric charge
(d) a stationary electric charge
[5] The circuit in the diagram at right will allow
what signals to pass?
(a) low frequencies (b) high frequencies
(c) DC signals
(d) no signals.
Coils do not allow the direction or
magnitude of the current to change
rapidly (Lenz’s Law) and thus block high frequencies.
[6] Which of the following is NOT a true statement about electromagnetism?
(a) A changing electric field generates a magnetic field.
(b) A changing magnetic field generates an electric field.
(c) A moving charge generates a magnetic field.
(d) A magnetic field in a closed iron loop generates an electric field.
Again, no changing magnetic flux. (a) is Maxwell’s displacement current.
[7] The Earth’s geographic south pole is a magnetic
(a) north pole.
(b) south pole.
(c) vortex.
(d) mess
[8] In an LC oscillator, it is NOT TRUE that
(a) the frequency is given by (L/C). Should be 1/(LC)
(b) energy is stored in the electric field of the capacitor.
(c) the current in the coil is at maximum when the charge on the
capacitor is zero.
(d) the magnetic field in the inductor keeps the current flowing when
the energy in the capacitor is zero.
[9] A charged particle enters a uniform magnetic field moving perpendicular to
the field lines. Which of the following is true? mv=Bqr
(a) Doubling the mass halves the radius of the circular path.
(b) Doubling the velocity has the same effect as doubling the charge.
(c) Doubling the charge gives one fourth the radius.
(d) A particle with the opposite charge will curve in the opposite direction.
[10] Two magnets repel each other. Their field lines in the region between them
(a) point in the same direction.
(b) point in opposite directions.
(c) must point away from the magnets.
(d) run from the pole of one magnet to the pole of the other.
[11] A transformer is to be used to decrease the voltage from the 110 V wall outlet. The
primary coil is plugged into the wall, the secondary provides the decreased voltage. The
number of turns in the secondary coil
(a) is less than the number of turns in the primary.
(b) is greater than the number of turns in the primary.
(c) may be greater or less, depending on the diameter of wire.
(d) does not determine the output voltage.
[12] Eyman Erdews Loh hooks the primary of a transformer to a 3V battery. The
secondary has twice as many turns as the primary. In his circuit, the voltage in the
secondary after a second or so will be
(a) 6V
(b) 1.5V (c) 0V (d) subject to the Ah-Fenner-Ong principle.
Again, no changing magnetic flux.
[13] The magnetic field produced by a toroid
(a) is a uniform field
(b) is four times that produced by a hemiroid.
(c) is zero outside the toroid.
(d) falls off as 1/r away from the toroid.
(d) is true inside the toroid.
[14] Miss Duppoint, in pondering her physics text, forms a question: If I enclose the
north pole of a magnet in a closed surface, what would be the total magnetic flux through
the surface? The answer to her question is
(a) positive. (b) negative. (c) zero.
(d) depends on the shape of the surface.
See question 2
[15] Coil A has one turn, Coil B has N turns of the same shape and area as coil A. This
will mean that, in similar situations, Coil B will generate
(a) N times the B magnetic field. (b) N times the induced voltage
(c) N times the inductance (d) all the above. (e) an N-oying answer.
Strictly speaking, (a) and (b) are true, (c) should be N2 times. I gave 2
points for (a) or (b).
Long answer. Do either A or B, Show all your work.
5 points each for parts a through e.
[A] A long solenoid of outside radius 30 cm. is wrapped with 1000 turns per meter. A
coil of 50 turns is wrapped tightly around outside of the solenoid at the center and is
connected to an ideal voltmeter. The solenoid wire carries a current of 20 mA.
(a) Calculate the magnetic flux through the outer coil.
Since the coil is wrapped tightly around the solenoid, the flux through the
coil is approximately the same as the flux through the solenoid. The
magnetic field inside a solenoid is uniform, so the flux is just the field times
the area:
B = o n I = (4 E-7Tm/A)(1E3/m)(2E-2A) = 2.5 E-5 T
 = BA = (2.5 E-5 T)(0.3m)2 = 7.1 E-6 Tm2
(b) Calculate the potential induced in the outer coil when a switch is opened and the
current through the solenoid drops to zero in 0.02 seconds. We are asking for magnitude
only here.
 = -N d/dt() = -50 /t = -50(7.1 E-6 Tm2)/(0.02 s)
= -17.7 mV (I didn’t worry about the sign)
(c) If the current in the solenoid wire was in a clockwise direction around the core, what
direction will the induced current in (b) be in? Before the switch is opened, what
direction is the force on the coils of the solenoid produced by the magnetic field inside?
Since the current is dropping, the induced current will try to maintain the
field (Lenz’s Law) and so it will be in a clockwise direction.
Before the switch is opened, the current is clockwise and so the B field is
down. IxB is therefore outward. While it is true that the field at the wire
is zero for a single layer solenoid, the outward force is a real problem for a
multilayer solenoid, which is more commonly the case.
(d) Calculate the inductance of the solenoid.
I had to give you a length of 0.5 meters in order to calculate this.
L = oN2A/ℓ = (4π E-7)(500)2π(0.3m)2 = 0.089 H
If you used a different length and the correct number of turns for that
length, I should have given you credit.
(e) What total energy was stored in the magnetic field of the solenoid before the switch
was opened?
U = ½ L I2 = ½ (0.089 H)(0.02A)2 = 8.9E-4 J
[B] An electron and a positron both moving at 5E4m/s in the +x direction enter a uniform
magnetic field of 0.2 Tesla in the +z direction.
(a) Draw a diagram of the motion of the two particles in the magnetic field.
The diagram should show the proper
direction of curvature for the particles
in the field.
(b) Calculate the cyclotron frequencies for the two particles.
ω = 2π v/2πr = v/r mv=Bqr;
v/r = Bq/m = ω = (0.2T)(1.6E-19 C)/(9.11E-31 kg) = 3.5E10 rad/s
(c) If the positron were an alpha particle instead, what would the radius of its path be?
m = 4(1.6E-27 kg), q = 3.2E-19 C r = mv/Bq
r = 4(1.6E-27 kg)(5E4 m/s)/(0.2T)(3.2E-19 C) = 5 mm
(d) Calculate the force exerted on the electron by the magnetic field.
F = qvB (perpendicular) = (-1.6E-19 C)(5E4 m/s)(0.2T) in the –ŷ direction
= 1.6E-15 N in the ŷ direction
(e) If the uniform field of 0.2 Tesla is produced by a solenoid carrying a current of 10
Amps, how many turns per meter must it have?
B = μo n I ; n = B/( μo I) = 0.2T/[(4πE-7 Tm/A)(10 A)] = 1.59E4 turns/m
Do either X or Y (20 points)
[X] A long wire and a circular coil of radius R with N turns lie next to each other in a
plane with the center of the coil distance d from the wire. Currents flow in such a
direction that the magnetic fields produced by the current in the wire and the current in
the coil are in the same direction at the center of the coil.
(1) Draw a diagram and indicate the direction of current flow in the wire and the loop.
(2) If both the coil and wire carry a current of I, derive an expression for the total
magnetic field at the center of the coil using superposition.
(1)
(2) B(straight wire) = μoI/(2πr)
B(coil) = N μoI/(2r)
If the currents are the same, then we just put in the correct distances and
add the two fields:
total B = μoI/(2πd) + N μoI/(2R)
[Y] Jupiter has a toroid of ions encircling it near the orbit of its moon Io. The ions have
charges of +e and are mostly oxygen and sulfur ions from Io’s volcanoes. Suppose there
are N such ions circling Jupiter at speed v and orbital radius R.
(a) Derive an expression for the magnetic field holding them in orbit (Jupiter’s magnetic
field).
mv = BeR so B = mv/eR with m = mass of oxygen or sulfur ion being
considered.
(b) These orbiting ions broadcast a signal in the radio wave region. Derive an expression
for the frequency of the particles’ orbit, (which equals the one frequency of radio
transmission).
This is the cyclotron frequency ω = 2πv/(2πr) = v/R
(c) Find an expression for the magnetic field generated by the orbiting particles.
The particles constitute a ring current, so the field at the center (center of
Jupiter) due to this current will be
B = μoI/(2R)
The current will be N e v/(2πR) C/s [ qv = Iℓ; I = qv/ℓ ]
This gives B = μo N e v/[(2πR) (2R)] = μo Nev/(4πR2)