PHY2049 Spring 2005
Prof. Darin Acosta
Prof. Paul Avery
April 4, 2005
PHY2049, Spring 2005
Exam 3 Solutions
1. Which of the following statements is true about the LR circuit shown?
(1) Just after the switch is closed R1 carries
zero current.
(2) Just after the switch is closed the two
resistors carry identical current.
(3) Just after the switch is closed the
current in R1 is greater than that in R2.
(4) A long time after the switch is closed
R1 carries zero current.
(5) If the switch is opened after being
closed for a long time the current in both
resistors immediately drops to zero.
It is (1):
Before the switch is closed, there is no current flowing through the inductor. When the
switch is closed, a current will try to start flowing through the inductor, but by Faraday’s
law and Lenz’s law, the inductor will set up an emf to resist the increasing B-field, thus
trying to keep the current zero.
Another way to see this is that immediately after the switch is closed, the voltage across
R2 is the same as that of the battery. So the current in the branch containing the inductor
is just that for an LR circuit containing only R1 and L. The expression for that is:
R
−t 1 
ε 
i1 = 1 − e L  = 0 when t = 0
R1 

2. Suppose that the battery in the previous problem is replaced with an AC source with
emf given by ε = 40sin (120π t ) . Which one of the following equations can possibly
describe the current in the inductor?
Again, the current in the inductor is given by that of a driven LR circuit containing only
R1 and L, since the voltage across R2 is given by the time-varying emf ε = 40sin (120π t ) .
The current in an inductor for a mostly inductive load (reactance from any capacitance is
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PHY2049 Spring 2005
XL
, so the only choice among the
R1
possibilities that shows this lag and is driven at the same frequency is:
negligible) always lags the voltage: tan φ =
i = 10sin (120π t − 0.4 )
Note that for this solution, the current peaks at a slightly later time than the voltage
because of the negative sign in front of the 0.4 phase constant.
3. An RLC series circuit has R=200Ω, L=0.1 mH and is driven by an AC source with emf
ε = 20sin (1800π t ) . For what capacitance will the current be in phase with the voltage
source?
The current will be in phase with the voltage when the circuit is driven at the natural
oscillation frequency:
1
ωL −
ωC = 0
tan φ =
R
1
⇒ ω2 =
LC
1
1
⇒C = 2 =
= 310µ F
2
ω L (1800π s −1 ) ( 0.1×10−3 H )
4. At what rate in V/sec must the potential difference between the plates of a parallelplate capacitor with a 300 µF capacitance be changed to produce a displacement
current of 0.18 A?
The appropriate law is Maxwell’s Law on induction:
dΦ
!∫C B ⋅ ds = µ0ε 0 dt E
dΦE
id ≡ ε 0
= displacement current
dt
Φ E ≡ !∫ E ⋅ dA = electric flux through plate of capacitor
S
To get the electric field from the voltage, one way is to note that the electric field of a
charged plate is given by:
E=
σ
q
CV
=
=
ε0 ε0 A ε0 A
From this we can solve for the displacement current:
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PHY2049 Spring 2005
dΦE
d  CV 
dV
A = C
= ε0 
dt
dt  ε 0 A 
dt
0.18 A
dV id
= =
= 600
dt C 3 × 10−4 F
id ≡ ε 0
5. A beam of unpolarized light moving downward along the +z axis is sent through a
system of three polarizing sheets whose polarizing directions make angles of θ1 = 30°,
θ2 = 50°, and θ3 = 90° to the y axis. What fraction of the light's initial intensity is
transmitted?
The intensity of unpolarized light that passes through one polaroid filter, no matter what
the angle, is always cut in half (the average of cos2θ is ½):
1
I1 = I 0 .
2
Once the light passes through that filter, its polarization is defined by the direction of the
filter. The intensity of light passing through the second filter is then given by:
I 2 = I1 cos 2 (θ 2 − θ1 ) = I1 cos 2 ( 20" ) ,
since it is the relative angle between the two filter sheets that matters. The intensity of
light coming from the third sheet is thus:
1
I 3 = I 2 cos 2 (θ3 − θ 2 ) = I 2 cos 2 ( 40" ) = I 0 cos 2 ( 20" ) cos 2 ( 40" ) = 0.26 I 0
2
Note that there is nothing special about the 90° angle that would make the intensity 0. It
would only happen if the middle sheet was aligned at 0°.
6. If the magnetic field in a plane electromagnetic wave is given by
B = Bm sin ( kx + ωt ) zˆ , in SI units, then the electric field is given by:
Note that this traveling wave moves in the −xˆ direction. For example, to keep the phase
constant inside the sine function, x must decrease for increasing t. We also know that the
1
E× B , which means that the electric field must
direction of the wave is given by S =
µ0
point in the −yˆ direction with the magnetic field in the ẑ direction. Finally, we know that
E
the ratio of the electric field to the magnetic field is the speed of light: m = c . Putting
Bm
all that together gives: E = −cBm sin ( kx + ωt ) yˆ
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PHY2049 Spring 2005
7. The diagram shows the passage of a ray of light from air into a substance X. The listed
angles are measured in degrees. The index of refraction of X is:
Snell’s Law applies to the angles with respect to the normal to the surface:
nair sin 40" = nX sin 20"
⇒ nX =
sin 40"
= 1.9
sin 20"
since the index of refraction of air is very nearly 1.0.
8. An electrically neutral object that is paramagnetic is brought toward one of the open
ends of a solenoid magnet (that is switched on). Which of the following is true:
(1) A force will pull the object into the solenoid
(2) A force will repel the object away from the solenoid
(3) No force is exerted on the object since it is neutral
(4) The object will levitate above the solenoid if the solenoid axis
is oriented vertically
(5) The object will exhibit hysteresis
An object that exhibits paramagnetism is one which develops magnetization in the same
direction as an applied magnetic field (the atomic magnetic dipoles line up). So if the
object is brought toward the open end of the solenoid, the magnetization is in the same
direction as the B field lines.
∂B
The force on such an object is given by: Fz = µ z
∂z
If the open end has B pointing in the z direction, and the B field is increasing in the z
direction (because the open end of the solenoid has a non-uniform field, with increasing
B toward the inside of the solenoid), the force is in the z direction. This means that the
object will tend to be pulled into the solenoid magnet where the field is increasing. The
answer is (1).
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PHY2049 Spring 2005
9. A rectangular loop of wire sits in the x-y plane in a region where there is a uniform
magnetic field of B = B0 zˆ , where B0 = 0.5 T. Two of the sides of the loop are parallel to
the y-axis and have a fixed length ly=5 cm. The remaining two sides are parallel to the xaxis and have a time-dependent length given by lx = a + vt, where a=5 cm and v=25
cm/s. What is the magnitude of the induced emf in the wire loop at time t=15 s?
The appropriate physical law is Faraday’s Law of induction:
dΦB
ε =−
dt
Φ B = B ⋅ A = magnetic flux
dA
d
= − B # x ( a + vt )  = − B# x v
dt
dt
| ε |= ( 0.5 T )( 0.05 m )( 0.25 m/s ) = 6.25 ×10−3 V
⇒ ε = −B
10. A charged capacitor is connected across an inductor to form an LC circuit. When the
charge on the capacitor is 0.25 C the current is 4 A. If the maximum current is 5 A, what
is the period of LC oscillations in seconds?
1 2 q12
Li1 +
2
2C
The subscript “1” denotes the values for the current and charge at the first time. At a
later time, all the energy is in the inductor (when there is maximum current), so:
1 2
U = Limax
2
Equating these energies gives us:
In an LC circuit, the total energy must remain constant: U =
1 2 q12 1 2
= Limax
Li1 +
2
2C 2
q2
2
⇒ imax
− i12 = 1
LC
U=
⇒ LC =
q12
1  T 
= 2 =

2
2
imax − i1 ω
 2π 
⇒ T = 2π
2
q12
= 0.52 s
2
imax
− i12
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