Download answer key 207 test iii spring 2013

Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
Sections 07 and 08
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USE THE COMPUTER BUBBLE SHEET TO ANSWER PROBLEM [I]
WHICH CONSISTS OF 12 MULTIPLE CHOICE QUESTIONS
AT 5 POINTS EACH FOR A TOTAL OF 60 POINTS
DO ONLY ONE OF THE OTHER TWO LONG PROBLEMS , [II] or [III]
WHICH IS WORTH 40 POINTS
CROSS OUT ONE OF THE BOXES [2] OR [3] ON THIS FRONT PAGE
TO INDICATE WHICH PROBLEM [II] OR [III] YOU WILL NOT DO
Your signature signifies that you will obey the HONOR CODE
You may be asked to show your photo ID during the exam.
PUT AN X NEXT TO YOUR DISCUSSION SECTION:
[ ] Dr. Mezincescu 5O, T 9:30 - 10:20 p.m.
[ ] Dr. Curtright 5P, T 11:00 - 11:50 p.m.
[ ] Dr. Huerta 5R, T 2:00 - 2 :50 p.m.
THE FORMULAE SHEET IS THE LAST PAGE.
You may detach it carefully.
Physics 207 TEST 3
April 15 2013
3
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
[I] This problem has 12 multiple choice questions. Choose the best answers.
REMEMBER TO BUBBLE IN THE ANSWERS IN THE COMPUTER
GRADED SHEET
[1.] In the figure to the left below the currents of magnitudes I1 , I3 , and I5 are flowing out of
Hthe paper, and the currents I2 and I4 are flowing into the paper as shown. The integral
~ · d~` in the counterclockwise direction is equal to
B
C
[A] µ0 (I1 + I2 + I3 ) [X] µ0 (I1 − I2 + I3 ) [C] µ0 (I1 − I2 − I3 + I4 + I5 )
[D] µ0 (−I1 + I2 + I3 − I4 − I5 ) [E] µ0 (I1 − I2 − I3 )
θ θ
H
i4
i5
H
i2
i1
⦿
I ⨂
wire segment of
length L with
a
current I into the
paper
i3
C
I wire segment of
length L with
current I out of the
paper
a
[2.] The figure to the right above shows two wire segments of length L carrying currents I into
and out of the paper. The wire segments each have a mass m and are suspended from
wires of length H in equilibrium against gravity by their magnetic forces. The angle θ is
related to the given quantities by the equation
[A] tan θ =
µ0 IL
2πa
[B] tan θ =
µ0 IL
4πamg
2
[X] tan θ =
µ0 I 2 L
4πamg
2
Lmg
µ0 I
[D] tan θ = µ0 I4πa
[E] tan θ = 4πamg
[3.] The figure to the left below shows a magnetic field increasing in time as Bz (t) = B0 +B1 Tt ,
where B0 , B1 , and T are all positive. The magnetic field is uniform in space with cylindrical
symmetry. The z axis points out of the paper. Use Faraday’s law to find the force on the
charge q located at rest at the point shown.
[A] F~ = qπr2 B1 ̂ [B] F~ = qrB1 ̂ [C]F~ = −qπr2 B1 ̂ [X] F~ = −qrB1 ̂ [E] F~ = qrB1 ı̂
T
2T
Bz
T
2T
Ez
y
z axis out of
the paper
r
q
y
z axis out of
the paper
r
x
EC
C
loop
2T
x
BC
C
loop
[4.] The figure to the right above shows an electric field decreasing in time Ez (t) = E0 − E1 Tt ,
where E0 , E1 , and T are all positive. The z axis points out of the paper. The electric
field is uniform in space with cylindrical symmetry . Use Ampere-Maxwell’s law to find
the induced magnetic field BC in the direction shown at a distance r from the center of
symmetry.
[A] BC =
[D] BC =
µ0 0 rE1
2T
µ0 0 E1
− T
Physics 207 TEST 3
[B] BC = −µ0 0 πr2 ET1
[X] BC =
[C] BC = µ0 0 πr2 ET1
−µ0 0 rE1
2T
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
Physics 207 TEST 3
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
[5.] The figure to the left below shows a metallic rod of length L moving at constant speed v
in the positive x direction inside a uniform magnetic field into the paper of magnitude B.
The emf developed in the rod is
[X] E = BLv ↑
[B] E = BLv ↓
[C] E = 0 [D] negative z
[E] negative x
y
⨂
⨂
L
⨂
⨂
The wire frame is moving toward the right
B
⨂
F
A
F
⨂
v
⨂
⨂
⨂
B
⨂
C
F
F
F
F
F
F=0
B
x
z
Inside the dashed square B is out of the paper. B=0 outside the square
[6.] The figure to the right above shows a metallic frame traveling through a region with a
uniform magnetic field pointing out of the paper. The frame is viewed at three different
times as it moves from left to right. For each instant an arrow representing the total
magnetic force on the frame is indicated. Which sequence (A, B, C) correctly describes
the magnetic force at each of those instants?
[A] A
[B] B
[X] C
[D] None of these
[7.] The RL circuit shown below has a battery with emf E. The switch has been open for a
long time and is then closed at t = 0. Immediately after the switch is closed, what is the
current i(t = 0+ ) through the battery and the emf EL (t = 0+ ) across the inductor?
2E
2E
[A] i(t = 0+ ) = 3R
, EL (t = 0+ ) = 0 [B] i(t = 0+ ) = 3R
, EL (t = 0+ ) = 2E
3
E
, EL (t = 0+ ) =
[X] i(t = 0+ ) = 2R
[E] i(t = 0+ ) = 0, EL (t = 0+ ) = E
E
2
[D] i(t = 0+ ) =
E
2R ,
EL (t = 0+ ) = 0
R
S
ℇ
L
R
+
i
R
-
[8.] In the circuit above the switch has been closed for a long time. A long time after after the
switch is closed find the current i(t = ∞) and the emf EL (t = ∞).
2E
2E
[X] i(t = ∞) = 3R
, EL (t = ∞) = 0 [B] i(t = ∞) = 3R
, EL (t = ∞) = 2E
3
E
[C] i(t = ∞) = 2R
, EL (t = ∞) =
[E] i(t = ∞) = 0, EL (t = ∞) = E
Physics 207 TEST 3
E
2
[D] i(t = ∞) =
E
2R ,
EL (t = ∞) = 0
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
Physics 207 TEST 3
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
[9.] If in the figure to the left below the the magnet is moving away from the coil,
[A] a current I is induced in the coil in the direction shown and the magnet is repelled
[X] a current I is induced in the coil in the direction shown and the magnet is attracted
[C] a current I is induced in the coil opposite to the direction shown and the magnet is
repelled
[D] a current I is induced in the coil opposite to the direction shown and the magnet is
attracted
[E] no current id induced in the coil
a
N
S
b
i(t)
v
I
L
[10.] In the figure to the right above the inductance L = 0.1 H. At the instant of time shown,
di
i = 10 A, and dt
= 100 A/s. The emf in the inductor produces a potential difference
Va − Vb given by
[A] 1 V [B] -1 V [C] -10 V [X] 10 V
[11.] An L-C circuit oscillates with a frequency f . If the capacitance and the inductance are
are reduced by a factor of 2 (L → L/2 and C → C/2) the frequency of oscillation of the
circuit
[A] remains the same [B] decreases by a factor of 2 [C] decreases by a factor of 4
[D] increases by a factor of 4 [X] increases by a factor of 2
[12.] In the figure below there is a magnetic field of magnitude B into the paper. The rod of
length L lies flat on the paper. It is moving with speed v as shown. The induced emf
produces a potential difference Va − Vb given by
[A] −BLV
[B] BLv
[C] −BLv sin θ
⨂
[X] BLv sin θ
⨂
B
a
⨂
⨂
[E] −BLv cos θ
⨂
L
θ
⨂
v
⨂
⨂
b
⨂
Physics 207 TEST 3
⨂
⨂
⨂
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
Physics 207 TEST 3
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
ANSWER ONLY PROBLEM II OR III
[II.] A square loop of wire with side length L and resistance R is moved across a uniform
magnetic field that exists only inside a square region of side 2L. The magnetic field of
magnitude B points out of the paper as shown in the figure. Let the coordinate x be the
position of the right side of the loop marked as side a-b.
[a] Graph the x component Fx of the external force need to move the loop at constant speed
v0 as a function of the coordinate x from x = −3L to x = 3L.
[b] For the case in [a] graph the induced current in the loop as a function of x x = −3L to
x = 3L. Take counterclockwise currents to be positive.
[c] Say that the side labeled a-b enters the B field at t = 0 with an initial speed v0 , and that
now there is no external force acting on the loop. Using Fx = max show that the velocity
as a function of time is vx (t) = v0 e−t/τ during the time interval when the side a-b moves
from x = −L to x = 0, and find τ .
[d] Find the position of the loop x(t) for the time interval 0 ≤ t ≤ tcenter where x(0) = −L,
and x(tcenter ) = 0, that is, do the integral
Z t
L
.
x(t) − (−L) =
v(t) dt while − L ≤ x ≤ 0, and show that e−tcenter /τ = 1 −
v0 τ
0
[e] Use the above to find the speed at the time tcenter , that is, v(tcenter ) in terms of v0 , L and
τ.
Answers:
[a] and [b]
2L
L
a
v
B
b
x
-3L
-L
-2L
0
L
2L
3L
Fx
B2L2v
R
[a]
x
-3L
-2L
0
-L
[b]
BLv
R
L
2L
3L
I
x
-3L
-2L
-L
0
L
2L
3L
[c] See next page
Physics 207 TEST 3
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
[c] While the side a-b moves from x = 0 as it enters the magnetic field, until it reaches x = L,
when the loop is completely inside the magnetic field, there is a force toward the left given
by
B 2 L2 vx
dvx
dvx
B 2 L 2 vx
dvx
B 2 L 2 vx
Fx = −
, but Fx = m
, so m
+
= 0, or
+
= 0.
R
dt
dt
R
dt
Rm
The solution of this differential equation with the initial condition v(t = 0) = v0 is
v(t) = v0 e−t/τ , where τ =
Rm
.
B 2 L2
[d]
Z
x(t) − (−L) =
t
Z
v(t) dt =
0
t
−t/τ
.
= v0 τ 1 − e
t
v0 e
−t/τ
dt = −v0 τ e
−t/τ 0
0
But x(tcenter ) = 0, therefore L = v0 τ 1 − e−tcenter /τ , or e−tcenter /τ = 1 −
L
.
v0 τ
[e] We have that
v(tcenter ) = v0 e
−tcenter /τ
Physics 207 TEST 3
, so using [d] we get v(tcenter ) = v0
L
1−
v0 τ
= v0 −
L
.
τ
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
[III.] (40points) 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 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] Say that the sliding wire has mass m and its initial velocity is vx (t = 0) = v0 . Show that
if there is no external force applied to the wire its velocity is given by vx (t) = v0 e−t/τ and
find τ in terms of m, R, B , and L.
[d] Find the total distance the wire moves before coming to rest , that is do the integral
Z ∞
x(∞) =
vx dt.
o
[f] Since the velocity depends on time, the current also depends on time. Calculate the total
heat energy dissipated in the resistor by calculating the integral
Z ∞
heat =
I 2 (t)R dt.
0
⨂
⨂
⨂
⨂
⨂
⨂
L
⨂
⨂
⨂
R
⨂
⨂
⨂ ⨂
⨂
⨂
⨂
⨂
⨂
vx
⨂
⨂
B
⨂
⨂
⨂
⨂
⨂
x
Answers:
[a] The emf is E = BLvx and the current flowing up in the wire is I =
BLvx
R .
[b] The magnetic force on the wire is toward the left, so Fx = −ILB = − B
2
2
L 2 vx
.
R
2
dvx
B L vx
x
[c] Using that Fx = m dv
= 0. The solution of this differential
dt , we get that m dt +
R
equation with initial condition v(t = 0) = v0 is v(t) = v0 e−t/τ , with τ = BRm
2 L2 .
∞
R∞
[d] x(∞) = 0 v0 e−t/τ dt = −v0 τ e−t/τ = v0 τ .
0
[e]
∞
B 2 L2 2 −2t/τ
B 2 L2 2 −τ −2t/τ heat =
I (t)R dt =
e
v e
dt =
v
R 0
R 0 2
0
0
0
0
2 2
2 2
B L 2 τ
B L 2 Rm
1
=
v0
=
v0
= mv02 .
2
2
R
2
R
2B L
2
So, the total energy dissipated in the resistor equals the initial kinetic energy of the sliding
wire.
Z
∞
2
Physics 207 TEST 3
Z
∞
B 2 L2 vx2
R dt =
R2
Z
∞
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
R
√ dx
x2 +a2
= ln(x +
e = 1.6 × 10−19 C,
√
x2
FORMULAE
SHEET TEST III
R
R
xdx
−1
dx
2
+ a ),
= (x2 +a
=
2 )1/2 ,
(x2 +a2 )3/2
(x2 +a2 )3/2
me = 9.1 × 10−31 kg,
mp = 1.67 × 10−27 kg,
k=
x
a2 (x2 +a2 )1/2
1
4π0
2
= 9 × 109 NC·m
2 .
Coulomb’s Law, Electric Fields (N/C), Electric Field Flux Φ, and Gauss’ Law
q1 q2 H
kQ
~q = q E,
~
~ Q | = kQ
~ · dA
~ = qinside
F~ =
F
|
E
,
V
=
,
Φ
=
E
2
2
closed
surface
4π0 r
r
r
0
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 uniform electric
~ are F~total = 0, U = −~
~ ~τ = p~ × E.
~
field E
p · E,
p
Harmonic Oscillator: Fx = −kx, ω = k/m
Electric Potential and Energy: Wagent = (K + U )f − (K + U )i , Wf ield = Ui − Uf
Rf
~ · d~s, E
~ = −∇V
~ = − ∂V ı̂ − ∂V ̂ − ∂V k̂, Er = − ∂V
Uq (~r) = qV (~r), Vi − Vf = i E
∂x
∂y
∂z
∂r
Potential energy U and work W to assemble configuration of charges:
X
qi qj
U =W =
4π0 rij
ij pairs
Capacitance Q = CV, U =
Q2
2C ,
0 E 2
κo A
2 , C =
d
1
,
parallel
Ceq
C2
uE =
series C1eq = C11 +
= C1 + C2
The voltage across a capacitor cannot change instantaneously.
2
2
q
Energy in a capacitor: UC = 2C
= CV2
R
~ σ = 1/ρ, V = iR, R = ρL/A
Circuits i = J~ · n̂ dA, J~ = q+ n+~vD+ + q− n−~vD− = σ E,
Resistors: in series Req = R1 + R2 , in parallel
1
Req
=
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.
R
~ × B, ~τ = µ
~ F~ = i ds
~ µ
Magnetic forces and Gauss’ law: F~ = q~v × B,
~ × B,
~ = I n̂A
H
~ · dA
~ = 0,
Gauss’, Biot-Savart, and Ampere-Maxwell laws:
B
I
N µ0 i
~ = µ0 iin = µ0 (ic + iD ), B = µ0 i for ∞ wire
~
~ · ds
dB =
d~s × ~r,
B
3
4πr
2πr
C
~ and
where the conduction current ic is given in terms of the conduction current density J,
the displacement current iD is given in terms of the displacement current density J~D , as
Z
Z
~
∂E
~
iin =
J · n̂ dA, and iD =
J~D · n̂ dA, with J~D = 0
.
∂t
S
S
Physics 207 TEST 3
April 15 2013
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 15 2013, 6:25 - 7:40 p. m.,
B
Faraday’s Law: E = − dΦ
dt
H
~ + ~v × B)
~ · d~`,
E = (E
E = −L ddti ,
1
E2 = −M di
dt
1
Mutual inductance: Φ2 = M i1 , E2 = −L di
dt .
di
Self inductance: ΦB = Li, E = −L dt
. A coil of length `, and cross sectional area A, with n turns
per unit length has a self inductance L = µ0 n2 `A.
The current in an inductor cannot change instantaneously
R-C, or R-L circuits: DC charging capacitor with q(t = 0) = 0 or inductor with iL (t = 0) = 0
q(t) = CEapplied (1 − e−t/τC ), τC = RC, iC (t) =
iL (t) =
dq
dt
Eapplied
(1 − e−t/τL ), τL = L/R.
R
Energy in an inductor: UL = 12 Li2 , the energy density in a B field is uB =
q
1
L-C circuit: ω = LC
Physics 207 TEST 3
B2
2µ0 .
April 15 2013