Comments in this font color are mine (i.e. – “Zane speaks”.)
[01] A device which converts electrical energy to mechanical energy is called a(n)
[A] generator [B] capacitor
[C] alternator
[D] motor
[E] miracle
[02] A 10 µF capacitor is connected in series with a 20 kΩ resistor across a 12- VDC source and
an open switch. If the switch is closed at t = 0 s, what is the charge on the capacitor at
t = 15 ms as a percent of the maximum charge, Q0?
[A] 0 C
[B] 192 %
[C] 7 %
[D] 37 %
[E] 45 %
q = Q0 (1−e−t/τ )
τ = RC = (20 x103 Ω)(10 x10−6 F) = 0.2s
⎡ −15x10−3 s ⎤
q
⎥ = 1−0.92 = 0.07=7%
= 1−exp ⎢
⎢ 0.2s ⎥
Q0
⎣
⎦
[03] Which of the following scenarios will NOT produce an electromagnetic wave?
[A] An AC current in a bare copper loop
[B] A charged rubber rod moving at constant velocity
[C] A charged rubber rod being waved up and down
[D] An electron traveling in a circle
[E] All of these will produce an EM wave
Only an ACCELERATING charged particle will ratiate an EM wave!
[04] A solenoid has N = 1353 turns, carries a current of 2.000 A and has a length of 34.00 cm.
The value of the magnetic field generated at the very center is
[A] 10.00 mT
[B] 3.400 mT
B = µ 0 nI
→n=
B = µ0 (
(
[C] 0.1000 mT
N
ℓ
[D] none of these
)
N
1353
)I = 4π x10#7 T • m / A
2.000 A ) = 10.00 mT
ℓ
(0.34 m) (
[05] A voltmeter measures the voltage of a household electrical outlet to be 120 VAC. This
value is the
[A] average voltage
[B] rms voltage
[C] peak voltage
[D] minimum voltage
[E] none of these
!
[06] An electromagnetic wave is propagating in the + ẑ -direction.
! If the E -field of this wave is
pointing in the ! ŷ -direction, the direction in which the B -field points must be
[A] + x̂
[B] + ŷ
[C] ! x̂
[D] ! ẑ
[E] none of these
[07] A very “tech-saavy” penguin is standing on an ice floe at the South Pole, holding a
compass. The needle of her compass points
[A] to the east
[B] to the west
[C] straight down into the ground
[D] straight up in the air
[E] none of these
The Earth’s magnetic field emerges from the South Pole and goes into the North Pole.
[08] Work done by the magnetic field on a charged particle is equal to (On Test 1 Spring 2015)
[A] qvB
[B]
mv
qB
! !
[C] q( v • B)
!
!
[D] B • ! s
[E] none of these
The magnetic force on a charged particle is perpendicular to its direction of motion and so
can do NO work.
[09] Dr.Bruce Banner requires exposure to gamma rays of a minimum frequency 8.5 x 1021 Hz
in order to become the Incredible Hulk. The wavelength of these gamma rays is
13
[A] 2.8x10 m
[B] 3.5x10
[D] frequencies this high do not exist
!14
m
[C] 3.5 µm
[E] none of these
λf = c
λ=
3x108 m / s
= 3.5x10"14 m
21
8.5x10 Hz
[10] A certain device requires 20 VAC for safe operation. It is plugged into a household outlet
Np
Ns
=
(120 VAC). The turns ratio,
primary
of the adaptor’s internal transformer is
secondary
[A] 0.17
[C] 6
Vp
Vs
=
[B] 60
[D] none of these
120 VAC
=6
20 VAC
[11] A household clothes dryer is plugged into an outlet designated as “220 VAC”. The
maximum voltage in the outlet is
[A] 311 VAC
[B] 220 VAC
[C] 156 VAC
[D] none of these
Vmax = 2Vrms = 2 (220 VAC) = 311VAC
[12]
Suppose we have a long narrow coil and wish to decrease its inductance. That can be
accomplished by
[A] decreasing the length
[B] increasing the current
[C] decreasing the cross-sectional area
[D] increasing the number of turns
[E] none of these
L=
µo N2A
→ As A decreases, L will decrease.
ℓ
[13] The plug-in GFCI (Ground Fault Circuit Interrupter) into which your hair dryer is plugged
keeps “tripping”. The most sensible thing for you to do is
[A] pull the GFCI out of the outlet and plug your hair dryer in directly.
[B] take your hair dryer to a different outlet with a different GFCI and try it there.
[C] plug a different device into the existing GFCI to see if it trips.
[D] B and C are both good ideas
[E] take your hair dryer outside and beat it with a hammer.
[A] is a VERY BAD idea! If your dryer is faulty, you could get a fatal shock!
[B] – If your dryer is faulty, you will trip the 2nd GFCI. If not, the original GFCI is bad.
[C] – if the 2nd device works OK, then you know your dryer has a problem. If not, the
GFCI needs to be replaced.
[D] – This is the correct answer. Both [B] and [C] will give you valuable information. This
process is called “trouble-shooting”.
[E] – While emotionally satisfying, this choice yields no usable information!
[14] An unpolarized beam of light of intensity I0 passes through two successive polarizers. The
axis of the 1st polarizer is at an angle of 20 ° to the vertical and that of the 2nd is at an
angle of 65 ° to the vertical. The intensity of the light emerging from the 2nd polarizer is
[A]
I0
4
[B] 0.44 I0
[E] none of these
I
I1 = 0
2
[C]
2I0
4
[D] zero
2
⎛ I ⎞⎟ 2
⎛ I ⎞⎟⎛⎜ 2 ⎞⎟
I
→I 2 = ⎜⎜ 0 ⎟⎟ cos (65°− 20°) = ⎜⎜ 0 ⎟⎟⎜⎜ ⎟⎟ = 0
⎜⎝ 2 ⎟⎠⎜ 2 ⎟⎟
⎜⎝ 2 ⎟⎠
4
⎝ ⎠
[15] A 20.0-µC particle with a mass of 4.00 x 10-5 kg moves perpendicular to a 5.00-T magnetic
field in a circular path of radius 25.0 m. The speed of the particle is
[A] 0.016 m/s
[B] 62.5 m/s
[C] 0.5 m/s
[D] none of these
mv 2
= qvB
r
qBr (20 x10!6 C)(5.00T)(25.0 m)
v=
=
= 62.5m / s
m
4.00 x10!5 kg
[16] A tee-shirt, which has printing on the front, appears to be black and white while indoors.
However, when the shirt is viewed outdoors in natural sunlight, vivid colors appear on it.
This is an example of
[A] electromagnetic induction
[C] magic
[D] polarization
[B] the photoelectric effect
[D] the photochromic effect
[17] The circuit pictured below has an AC source whose frequency is variable.
[A]
[B]
[C]
[D]
Calculate the impedance for the circuit as depicted (f = 60 Hz). (8 points)
Calculate the resonant frequency for the circuit. (6 points)
Calculate Imax for this circuit. (2 points)
What is the impedance of this circuit in the high-frequency limit (f → ∞)? What is the
current in the circuit for this frequency? (2 points)
[E] What is the impedance of this circuit in the low-frequency limit (f → 0)? What is the
current in the circuit for this frequency? (2 points)
⎛
⎞⎟2
1
[A] Z = R + (X L − X C ) = R + ⎜⎜ωL −
⎟
⎜⎝
ωC ⎟⎟⎠
2
2
2
⎛
⎜
1
2
Z = (175Ω) + ⎜⎜⎜2π(60 Hz) 90 x10−3 H −
⎜⎜⎝
2π(60 Hz) 15.0 x10−6 F
(
)
[B] At resonance, X L = X C , and Z = R
f res =
1
2π LC
[C] I max =
=
(
→ 2πfL =
1
−3
−6
2π (90 x10 H)(15x10 F)
2 ( Vrms )
Zat res
=
2 ( Vrms )
R
=
1
2πfC
= 137 Hz
2 (120 V )
175Ω
)
⎞⎟2
⎟⎟ = 226Ω
⎟⎟
⎟⎠
= 0.97 A
This is a series circuit!
[D] When f → ∞, X L → ∞, so Z must also → ∞! If →∞ = 0
[E] When f → 0, X C → ∞, so Z must also → ∞! If →0 = 0
[18] The figure shows a zero-resistance rod sliding to the right on two zero-resistance rails
separated by the distance L = 0.50 m. The rails are connected by a 100.0 Ω resistor, and the
entire system is in a uniform magnetic field with a magnitude of 0.90 T.
[A] Find the speed at which the rod must be moved to produce a current of 0.250 A in the
resistor.
[B] Find the force that must be exerted on the rod in order to maintain this current.
[C] Find the power dissipated by the resistor.
[D] Find the mechanical power delivered to the rod.
(5 points for each part)
[B]
E = IR = vB l
IR (0.250 A)(100.0 ΩΩ)
v=
=
= 55.55 m / s ≈⋲ 56 m / s
Bl
(0.90 T)(0.50 m)
F = I l Bsin θѳ = (0.250 A)(0.50 m)(0.90 T)(1) = 0.1125 N ≈⋲ 0.11 N
[C]
[D]
P = I 2 R = (0.250 A) 2 (100 ΩΩ) = 6.25 W
P = Fv = (0.11N)(56 m / s) = 6.25 W (Use all the decimal places!)
[A}