Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
Section 01I, MTWRF 10:05 – 11:30 p. m.
Name:
Signature:
1
ID number:
2
ENTER YOUR NAME AND FORM 1 (FM1)
IN THE SCANTRON SHEET
3
DO QUESTION [I] WHICH CONSISTS OF
12 MULTIPLE CHOICE PROBLEMS FOR A TOTAL OF 60 POINTS
DO ONE OF THE TWO LONG PROBLEMS PROBLEMS [II], OR [III]
FOR 40 POINTS
IF YOU DO BOTH LONG PROBLEMS
THE LAST ONE WILL BE IGNORED
CROSS OUT THE BOX 2 OR 3 OF THE PROBLEM YOU OMIT
Your signature signifies that you will obey the HONOR CODE
You may be asked to show your photo ID during the exam.
NO CELL PHONES, NO CALCULATORS, NO PENS
ONLY USE A PENCIL. NO BACKPACKS
THERE IS A SEPARATE FORMULA SHEET
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
[I] This problem has 12 multiple choice questions. Correct choices are marked
with [X]
[I.1] A cylindrical wire has a resistance R. A wire of the same resistivity with
half the length and half the diameter, will have a resistance given by
[A] 4R [X] 2R [C] R [D] R/2 [E] R/4
[I.2] The oscillations of a magnetic needle in a magnetic field are important in
explaining
[A] Ampere’s law [B] Faraday’s law [X] MRI
[D] eddy currents [E] transformers
[I.3] The figure to the left below shows two very long, straight, insulated wires
oriented at right angles. The wires carry currents of equal magnitude I in
the directions shown. The net magnetic field at point P is
µ0 I
µ0 I
0I
0I
[X] 0 [B] − µπr
ẑ [C] µπr
ẑ [D] 2πa
(x̂ + ŷ) [E] − 2πa
(x̂ + ŷ)
y
z
I
x
a
P
a
B
v
q
x
I
screen
[I.4] The figure to the right above shows a mass spectrograph. A particle with
mass m, charge q, and velocity v, enters a region of uniform magnetic field
B. It strikes the wall at a distance x from the entrance slit. If the particle’s
velocity stays the same, but its charge-to-mass q/m ratio is doubled, xnew
√
[A] xnew → 2x [B] xnew → 2x [C] x [D] xnew → √x2 [X] xnew → x2
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
[I.5] In the RC circuit shown to the left below the capacitor is initially uncharged.
The switch is closed at t = 0. Immediately after the switch is closed, what is
the current i2 (t = 0+ )? You do not have to solve any differential equations
to get the answer.
[A] i2 = 0 [B] i2 = 2 A
[C] i2 = 1.2 A
[X] i2 = 1.5 A
[E] i2 = 3 A
[I.6] In the same RC circuit below, as t → ∞ find the voltage VC across the
capacitor.
[A] VC = 0 [X] VC = 8 V [C] VC = 12 V [D] VC = 6 V [E] VC = 4.8 V
y
i1
12 V
+
−
B
i3
2Ω
I
i2
R
y
4Ω
4Ω
C
x
z
I
v
S
N
I
[I.7] The figure at the center above shows a semicircular loop of radius R that
lies flat on the paper. It carries a clockwise current I. The loop is immersed
~ = −B ẑ. Find
in a uniform magnetic field that points into the paper , so B
the total magnetic force on the loop using ı̂, ̂, k̂ notation.
~ = 2IRB ŷ [B] F
~ = −2πIRB ŷ [C] F
~ = 2πIRB ŷ
[X] F
~ = −2IRB ŷ [E] F
~ = 2IRB x̂
[D] F
[I.8] In the figure to the right above the coil has no battery and an induced
current I
[A] I is in the direction shown and the magnet is attracted
[B] I is in the direction shown and the magnet is repelled
[X] I is in the opposite direction to that shown and the magnet is repelled
[D] I is in the opposite direction to that shown and the magnet is attracted
[E] no current is induced in the coil
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
[I.9] The figure to the left below shows a time dependent magnetic field increasing
as Bz (t) = B0 + B1 Tt , where B0 , B1 , and T are all positive. The magnetic
field is uniform in space with cylindrical symmetry. Find the force on the
charge q located at rest at the point shown.
[A] F~ = qπr2 B1 ŷ [B] F~ = −qπr2 B1 ŷ [C] F~ = qrB1 ŷ
T
[X] F~ =
T
−qrB1
2T ŷ
2T
[E] F~ = 0
Bz
z axis out of
the paper
r
q
EC
i2
i1
x
z
i4
i5
y
i3
loop
C
C
[I.10] In the figure to the right above the currents of magnitudes I1 , I4 , and I5
are flowing out of the paper, and
H the currents I2 and I3 are flowing into the
~ · d~` is equal to
paper as shown. The integral C B
[A] µ0 (I1 −I2 +I3 −I4 +I5 ) [B] µ0 (I1 +I2 +I3 +I4 +I5 ) [C] µ0 (−I1 +I2 −I3 )
[D] µ0 (I1 − I2 + I3 )
[X] µ0 (I1 − I2 − I3 )
[I.11] The figure below shows 5 resistors to the left and 4 capacitors to the
right. Find the equivalent resistance Req between points A and B, and
the equivalent capacitance Ceq between points A and B.
C
R
A
R
A
R
R
C
B
R
C
B
C
[A] Req = R/2, Ceq = 5C/2 [B] Req = 5R, Ceq = 4C
[X] Req = R/2, Ceq = 3C/5 [D] Req = R/2, Ceq = 2C/5
[E] Req = R/2, Ceq = 5C/3
[I.12] In class a pendulum and magnet were used to demonstrate
[A] Ampere’s law [B] Gauss’ law [C] MRI
[X] eddy currents [E] transformers
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
[II.] (40 points) In the figure the wire of length L is sliding with velocity vx > 0
without friction on rails that form part of a circuit with resistance R. A
uniform magnetic field of magnitude B points into of the paper.
[a] Find the induced emf E and the current I in the circuit, indicating the true
positive direction of the current that flows in the sliding wire (up or down
on the paper).
[b] Find the magnitude and direction of the magnetic force on the sliding wire
in terms of B, L, vx , and R.
[c] The wire has an initial velocity vx (t = 0) = v0 , a mass m, and the only
force acting on it is the force found in [b]. Show that the velocity obeys the
equation
dvx
vx
+
= 0,
dt
τ
whose solution is vx (t) = Ae−t/τ ,
and find τ in terms of m, R, L and B. Also find the constant A.
[d] Since the velocity is a function of time, write the true current as a function
of time, and find the heat dissipated in the resistor from t = 0 to t → ∞,
for which you must do the integral
Z ∞
Heat =
i2 (t)R dt,
0
and interpret how this is related to energy conservation.
⨂
⨂
⨂
⨂
⨂
⨂
L
⨂
⨂
⨂
R
⨂
⨂
⨂ ⨂
⨂
⨂
⨂
⨂
⨂
vx
⨂
⨂
⨂
B
⨂
⨂
⨂
⨂
x
Answers: Take vx > 0 so the wire really is moving toward the right.
[a] For a Faraday loop we use a curve that surrounds the rectangular area
between the sliding wire and the resistor oriented in the clockwise direction.
Then in the rectangular area we use a normal n̂ into of the paper in the
~ This means that The magnetic flux is
same direction as B.
Z
Z
Z
~ · n̂ dA = B dA = B dA = BLx.
ΦB = B
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
The emf in the clockwise direction is
Eclockwise = −
dΦB
dx
dx
= −BL
= −BLvx < 0, because here
= vx .
dt
dt
dt
The clockwise current iclockwise = Eclockwise /R is negative, therefore icounterclockwise
is positive,
icounterclockwise =
BLvx
BLvx
, so iupward
=
.
wire
R
R
[b] The magnetic force on the wire is
(LB)2 vx
upward ~
upward
~
~
Fwire = iwire L × B = −iwire LB x̂ = −
x̂,
R
that is, toward the left in the figure, opposite to the direction of motion of
the wire.
[c] The force on the wire causes an acceleration
d~v
dvx
(LB)2 vx
dvx
F~wire = m , so Fx = m
, or −
=m
dt
dt
R
dt
therefore
m
(LB)2 vx
dvx
vx
mR
dvx
+
= 0, or
+
= 0 where τ = 2 2 .
dt
R
dt
τ
L B
The solution is
vx = Ae−tτ , but vx (t = 0) = v0 , therefore A = v0 , and finally vx (t) = v0 e−tτ .
[d] The current iR in the resistor is downward in the picture, and
Z ∞ 2 2
BLvx (t)
BL
B L
−tτ
iR (t) =
=
v0 e , therefore Heat =
Rv0 e−2tτ dt,
2
R
R
R
0
B 2 L2 τ 1
B 2 L2
−τ −2tτ ∞
=
so Heat = 2 R
v0 e
R(
= mv02 ,
2
0
R
2
R
2
2
so the heat dissipated in the resitor equals the initial kinetic energy of the
moving wire.
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
[III] (40 points) A generator consists of a wire loop of zero resistance rotating
with a positive angular frequency ω = dθ/dt, so θ = ωt, as shown. Sides
ad and bc have length H, and sides ab and dc have length L. There is a
magnetic field of magnitude B through the loop as shown. The rings #1
and #2 do not move. Wire ad makes sliding contact with ring #1 and wire
bc makes sliding contact with ring #2. The bottom figure shows the loop
viewed from the left, with an end view of wires ad and bc.
[a] The normal to the loop n̂ is defined as shown. Find the magnetic flux
through the loop.
[b] Use Faraday’s Law to find the emf EC in the loop where the orientation of
the loop C is shown.
[c] If the switch is closed find the current and at the moment shown indicate
its true direction upward or downward through the resistor R.
[d] At the moment shown find the torque on the loop and indicate its direction.
switch
B
θ
L
R
d
a
sliding contact
n
ω
2
1
b
C
ad
c
B
ω
θ
n
H
bc
Answers
[a] With a uniform magnetic field, and with θ = ωt, the magnetic flux through
the loop is
Z
Z
~ · n̂ dA =
ΦB =
B
B cos θ dA = B cos θA = BHL cos ωt
S
S
[b]
EC = −
dΦB
d
= −BHL (cos ωt) = BHL ω sin ωt
dt
dt
[c] The current is iC (t) = EC (t)/R upward through the resistor.
~ = µB sin θ ←, toward the left, so it opposes
[d] With µ
~ = iC HL n̂, ~τ = µ
~ ×B
the rotation, and ~τ = ω(BHL sin ωt)2 ←.
Physics 207 TEST 2 FORM 1 ANSWER KEY
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
FORMULAE SHEET TEST 2
me = 9.1 × 10−31
g = 10 m/s2 , e = 1.6 × 10−19 C
1
9 N ·m2
kg, mp = 1.67 × 10−27 kg, k = 4π
=
9
×
10
C2 .
0
Coulomb’s Law, Electric Fields (N/C)
q1 q2 ~
~Q =
F~ =
F~q = q E,
E
4π0 r 2
kQ
r 2 r̂,
Electric Field Flux Φ, and Gauss’ Law
I
Φclosed
surface
with qinside
~ · n̂ dA = qinside
E
0
ZS
Z
Z
=
ρ dV, or qinside =
σ dA, or qinside =
λ dl
=
V
A
Spherical symmetry : Er =
qinside
,
4πo r2
L
Cylindrical symmetry : Er =
λinside
2π0 r
Electric Field and sheets of surface charge density σ:
On each side of an infinite sheet En = 2σ0 . However, outside the surface of
a conductor, En = σ0 .
Force, Potential Energy and Torque for an Electric Dipole
The Force, torque, and potential energy of an electric dipole p~ immersed in
~ are F~total = 0, U = −~
~ ~τ = p~ × E.
~
uniform electric field E
p · E,
p
d2 x
2
Harmonic Oscillator: Fx = −kx, dt2 + ω x = 0, ω = k/m
agent
f ield
Work-Energy Theorem: Wi→f
= (K + U )f − (K + U )i , Wi→f
= Ui − Uf
Electric Potential and Energy:
Rf
Q
~ · d~s,
VQ (r) = 4π
,
U
(~
r
)
=
qV
(~
r
),
V
−
V
=
E
q
i
f
i
0r
~ = − ∂V ı̂ −
E
∂x
∂V
∂y
̂ −
∂V
∂z
k̂, Er = − ∂V
∂r ,
when E field is uniform E =
Physics 207 TEST 2 FORM 1 ANSWER KEY
V
d
June 17 2014
Dr. Huerta Phy 207 1st Summer Session Test 2 FORM 1 ANSWER KEY june 17 2014
Capacitance Q = CV , Energy in capacitor U =
CV 2
2
κo A
1
1
1
, series
=
+
, parallel Ceq = C1 + C2
d
Ceq
C1
C2
R
R
Current i = S J~ · n̂ = S Jz (r)2π rdr dA, J~ = q+ n+~vD+ + q− n−~vD− =
C=
Resistors: V = iR, R =
ρL
A ,
in series Req = R1 + R2 , in parallel
1
Req
=
~
E
ρ
1
R1
+
1
R2
Kirchhoff ’s rules: (1) at a junction, current in equals current out
(2) sum of voltage rises around a loop equals zero.
RC circuits
q
Differential equation: dq
solution q(t) = Ae−t/τ + Kτ
dt + τ = K,
−q0 −t/τ CE −t/τ
dq
−t/τ
−t/τ
=
e
+
e
, with τ = RC.
q(t) = q0 e
+CE 1−e
, i(t) =
dt
τ
τ
The voltage across a capacitor cannot change instantaneously.
Magnetic forces and Gauss’ law:
R
~ F~ = i d~` × B, ~τ = µ
~ µ
F~ = q~v × B,
~ × B,
~ = I n̂A
H
~ · dA
~ = 0, dA
~ = n̂ dA
Gauss’, Biot-Savart, and Ampere-Maxwell laws: B
~ = N µ0 i d~` × ~r,
dB
4πr3
I
~ · d~` = µ0 iin = µ0 (iC + iD ), B = µ0 i for ∞ wire
B
2πr
C
where the conduction current iC is given in terms of the conduction current
~ and the displacement current iD is given in terms of the displacement
density J,
current density J~D , as
Z
Z
~
∂E
~
~
~
iC =
J · n̂ dA, and iD =
JD · n̂ dA, with JD = 0
.
∂t
S
S
Faraday’s Law
R
~ · n̂ dA,
ΦB = B
R ∂~B~
R
dΦB
~ · dA~
E
=
−
=
·
n̂
dA
+
B
C
dt
dt
S
S ∂t
S
H
~ + ~v × B)
~ · d~`,
where E = C (E
H
R ∂B
R
~ · n̂ dA and H (~v × B)
~ · d~` = −
~ · d~` = B
~·
with C E
S ∂t
C
S
Physics 207 TEST 2 FORM 1 ANSWER KEY
~
dA
dt
June 17 2014