Exam III
PHY 2049
Summer C
July 16, 2008
1. In the circuit shown, L = 56 mH, R = 4.6 Ω and V = 12.0 V. The switch S has been
open for a long time then is suddenly closed at t = 0. At what value of t (in msec) will the current
in the inductor reach 1.1 A? (Figure, problem 1 PHY2054exam III, Spring 2008).
(1) 6.67
(2) 10.5
(3) 2.88
(4) 19.0
(5) None of these
The battery is connected directly across the series combination of the inductor L and resistor R.
Therefore the usual LR circuit formulas apply, even though the battery is also connected directly
across another resistor R. So the current in the inductor is given by:
i (t ) = i0 (1 − e − t /τ )
τ = L/R
i0 = V / R
Solving for time we have,
 i (t ) 
L 
i (t ) 
56mH 
1.1A 
t = − τ ln  1 −
= −
ln  1 −
= 6.67ms
 = − ln  1 −

i0 
R  V / R
4.6Ω
 12 / 4.6Ω 

2. Refer to the previous problem. What is the total energy stored in the inductor a long time after
the switch is closed?
(1) 0.19 J
(2) 0.048 J
(3) 0.76 J
(4) 0.034 J
(5) None of these
After a long time, the current i(t) → i0, as we can see from the formula above. So therefore the
energy is:
2
1
1
 12 
U = Li02 = (56 × 10− 3 ) 
 = 190mJ
2
2
 4.6 
3. A real inductor has resistance because it is composed of a coil of wire with nonzero
resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery
and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz
generator and measures an rms current of 0.86 A. What is the inductance of the coil?
(1) 0.081 H
(2) 0.51 H
(3) 34.9 H
(4) 30.6 H
(5) None of these
The coil has an inductance and resistance. We can visualize the coil therefore as an inductor
and resistor in series. When we connect the DC battery across the coil, we have therefore and
LR circuit. The steady state current is just i0 = V/R – same as above. So
V
12
=
= 16.67Ω
i0 0.72
R=
When we connect the AC battery across the coil, then we have:
irms =
Vrms
=
Z
30
16.67 + (2π ⋅ 60 ⋅ L) 2
2
= 0.86
The last gives,
L=
1
120π
900
− 16.67 2 = 0.08H
0.862
4. Sunlight of intensity 1340 W/m2 is incident normally on a 9 km × 9 km rectangular sail of
negligible mass attached to a spaceship of mass 12,000 kg. If the sail
is perfectly black, how long will it take the spaceship to move 1 km starting from rest?
(1) None of these
(2) 180 sec
(3) 8.1 sec
(4) 24 min
(5) 4.8 hr
The radiation produces a pressure on the sail, which is: P = I / c . This gives rise to a force
which accelerates the sail (spaceship). Therefore, using simple kinematics equation, and
Newton’s 2nd Law, we have:
∆x=
1 2 1F 2
at =
t →
2
2m
t=
2m∆ x
=
F
2m∆ x
=
PA
2mc∆ x
=
IA
2(12000)(3 × 108 )(1000)
= 257s
(1340)(81× 106 )
5. A dimmer for a stage light at a theater consists of a variable inductor, whose inductance is
adjustable between zero and Lmax, connected in series with a lightbulb. The electrical supply is
120 V (rms) at 60.0 Hz; the lightbulb is rated at 1.0 kW for this voltage. What Lmax is required if
you want the rate of energy dissipation in the lightbulb to be variable by a factor of 10 from its
upper limit of 1.0 kW? Assume that the resistance of the lightbulb does not depend on its
temperature.
(1) 0.11 H
(2) 0.45 H
(3) 1.2 H
(4) 65 mH
(5) 21 mH
First we’ll determine the resistance of the lightbulb. The problem states that without the
inductor, the average power dissipated by the resistor is 1kW. Therefore,
2
Vrms
→
R
V2
1202
R = rms =
= 14.4Ω
P 1000
P=
If we put the inductor in, then the current will decrease, and the power dissipated across the
resistor will therefore decrease. So want to know what Lmax will give rate of energy dissipation
equal to 1.0kW/10 = 100W. So then when add inductor at its maximum value, the power
dissipated across the resistor will be:
2

 Vrms 

P= 
R
=


 Z 

2
rms
2
Vrms
R 2 + ω 2 L2max
V R
= R 2 + ω 2 L2max → Lmax =
P
Lmax = 0.11H

V2 R
 R = 2 rms 2 2 →

R + ω Lmax

2
Vrms
R
− R2
P
→
ω2
6. A small light bulb is placed 3 cm inside a solid cube of Lucite (n=1.50) and turned on. What is
the radius (in cm) of the illuminated circle that would be seen by an observer standing above the
Lucite block?
(1) 2.68
(2) 3.00
(3) 3.35
(4) 4.50
(5) 3.42
The radius of the circle seen is the distance between the lightbulb and the point on the interface
where light at the critical angle would hit.
Therefore,
θ c = sin − 1 (n2 / n1 ) = sin − 1 (1/1.5) = 41.8
r = 3cm × tan 41.8 = 2.68cm
7. A horizontal beam of unpolarized light is incident upon a stack of 4 polarizers with axes of
polarization, in order and measured clockwise from the vertical, at 30°, 75°, 120° and 180°.
What is the ratio of the intensity of the transmitted beam to that of the incident beam?
(1) 0.031
(2) 0.063
(3) 0.023
(4) 0.047
(5) 0.0063
Since the initial light is unpolarized, half the light would get through the first polarizer. The
fraction that would get through the second is cos2α, where α is the angle between the incomming
light and the polarization axis of the polarizer. Since the light coming out of the 30 deg.
polarizer is now polarized at an angle of 30 deg., α would be 45 for the second polarizer. And
so on…so we end up with.
1

I 4 =  cos 2 45cos 2 45cos 2 60  I1 = 0.031I1
2

8. An oscillating LC circuit consists of a 2 mH inductive coil and a 4μF capacitor. The
capacitor has a potential difference of 0.75 V when the current through the coil is 30 mA.
What is the maximum possible charge on the capacitor [or current through the inductor]?
(1) 4 x 10-6C
(2) 3 x 10-6 C
(3) 2 x 10-6 C
(4) 8 x 10-9 C
(5) 0.75 C
We can use conservation of energy. The energy stored in the circuit is:
1
1
1
1 Q(t ) 2
LI (t ) 2 + CV (t ) 2 = LI (t ) 2 +
2
2
2
2 C
1
1
= (2mH)(30mA) 2 + (4μF)(0.75)2 = 2.02μJ
2
2
U=
and this is constant for all times. Now consider the particular time when Q is at its maximum
value, (and I is 0 therefore). Then we can also say
2
1
1 Qmax
L ⋅ 02 +
2
2 C
U=
2
1 Qmax
= 2.02μJ
2 (4μF)
Q max =
2.02μJ × 2 × (4μF) = 4.04μC
9. The average intensity of light from an incandescent light bulb is 300 mW/m on a particular
surface. Assuming that the light is in the form of an electromagnetic plane wave, what is the
maximum magnetic field amplitude, Bm?
2
(1) 5 x 10-8 T
(2) 7 x 10-8 T
(3) 3.5 x 10-8 T
(4) 21 T
(5) 15 T
I=
1 2
1 2
Brms c =
Bm c →
µ0
2µ 0
Bm =
2µ 0 I
=
c
2(4π × 10− 7 )(300 × 10− 3 )
= 50nT
3 × 108
where we use the relationship between Brms and Bm, namely Brms = Bm/√2.
10. Light traveling horizontally enters a right prism through the hypotenuse, as shown in the
figure. The index of refraction of the prism is n=1.6. At what angle is the light deflected from
horizontal?
(1) 36º
(2) 26º
(3) 19º
(4) 45º
(5) 31º
Snell’s law for the first interface is:
θ 2 = sin − 1
n1
1
sin θ 1 = sin − 1
sin 50 = 28.6
n2
1.6
Angle with top diagonal surface is 90 – 28.6 = 61.4. And at corner is 50. Therefore angle of
ray 2 with the vertical side of the triangle is 180 – 61.4 – 50 = 68.6. Therefore the angle of ray
2 with respect to the normal to the vertical side is 90 – 68.6 = 21.4. Snell’s law then gives,
θ 2 = sin − 1
1.6
sin 21.4 = 35.7
1
11. An arrangement for generating a traveling electromagnetic wave in the shortwave radio
region of the spectrum works as follows: an LC oscillator produces a sinusoidal current in the
antenna, which generates the wave, traveling outward at the speed of light. What is the
wavelength (in meters) of the wave emitted by this system if L = 0.323μH and
C = 45.0pF?
(1) 7.19 m
(2) 1.14 m
(3) 719 m
(4) 114 m
(5) None of these
The LC circuit’s current will oscillate at ω = 1/√(LC). And this will be the (angular) frequency
of the EM wave as well. From the frequency we can obtain the wavelength. So,
f =
λ =
1
2π
LC
=
2π
1
= 41.7MHz
(0.323µ H )(45 pF )
c
3 × 108
=
= 7.2m
f 41.7 × 106
12. A current of 1A is used to charge a parallel plate capacitor with square plates. If the area of
each plate is 0.6m2 the displacement current through a 0.3m2 area wholly between the capacitor
plates and parallel to them is:
(1) 0.5A
(2) 1A (3) 2A (4) 0.7A
(5) 0.25A
The displacement current through a loop of area A is defined as:
id = ε 0
dΦ E
d
= ε 0 ∫ E ⋅ dA
dt
dt
Our loop is inside the capicator and has an area A of 0.3m2. The Electric field, E, through the
loop is parallel to the area vector, A at all points. It is also constant over the surface of the
loop. But E does change with time since the charge on the plates (which is changing with time)
determines the field between the plates. Therefore we may write,
id = ε 0
d
d
d
d
EdA = ε 0 E ∫ dA = ε 0 EA = ε 0 A E
∫
dt
dt
dt
dt
To determine the rate of change of the electric field, we will relate the field between the plates,
to the potential difference between the plates, and then finally to the charge on the plates. So we
have,
E = V / d = Q / Cd
where Q is the charge on the plates, C the capacitance of the plates, and d the distance between
them. So now we have,
id = ε 0 A
d Q ε 0 A dQ ε 0 A
=
=
i
dt Cd Cd dt
Cd
Now let AC be the area of the plates of the capacitor. And then we’ll have,
id =
ε 0A
A
i=
i
AC
 ε 0 AC 
 d  d
So our answer is:
0.3m 2
id =
(1A) = 0.5 A
0.6m 2