Download Physics 213 Final Exam March 15, 2004

Physics 213
Final Exam
March 15, 2004
Please enter your last name and university ID number on front page of the answer sheets. Also
sign your name in the space provided at right. Before you begin the exam, fill in the ID
information requested on all pages of the solution papers.
Directions: The exam consists of 5 questions worth 40 points each. Only approved calculators
may be used for this exam. Palmtop or laptop computers are not allowed. A sheet of formulas
and constants is provided for your use. In solving problems, show your work and explain your
reasoning. Except for multiple-choice questions, no credit will be given for an answer that does
not include the necessary solution or explanation. Partial credit may be awarded for a correct
method of solution, even if the answer is wrong. Please place your answer to the multiple-choice
questions in the appropriate box on the grid provided(blacken the box with pencil). Work only on
the front of the pages of the exam booklet, and if more space is needed for your solution please
direct accordingly (i.e. continued on back page). Use only the exam booklet provided.
________________________________________________________________________
1. (4 points each) IT IS NOT NECESSARY TO SHOW YOUR WORK FOR THESE
MULTIPLE CHOICE QUESTIONS.
Choose the one correct answer for each of the following questions that best answers or completes
the question. Place your answer in the appropriate box on the grid provided for Question #1.
I.
Compared to the filament of a 100-watt light bulb, the filament of a 40-watt light
bulb (of the same material) is
A) thicker
C) same
B) thinner
D) not possible to determine without its resistivity.
2
Ans: B
P=V /R
2
V
V2
R100 
and R40 
100
40
l
l
R100 < R 40 and ρ
<ρ
Hence : A40 < A100 or r40  r100
A100
A40
II.
A moving electron can be accelerated without changing its speed by
A) an electric field .
D) This situation can not be possible.
B) a magnetic field.
E) none of these.
C) magnetic and electric field.
Ans: B
III.
Two loops of wires are side-by-side in the plane of the page. Initially no current flows in either loop. The
current in the left loop is gradually increased in CCW direction. Which of the following statements is
correct?
A) There is a current induced in the right loop, and it is in CCW direction.
B) There is a current induced in the right loop, and it is in CW direction.
C) There is no current induced in the right loop since the magnetic field produced by the current of the
left loop is entirely inside the loop.
D) There is no magnetic flux through the left loop.
E) There is no magnetic flux through the right loop.
Ans: A
Net B-flux is increasing in right loop. By lenz's
law the induced current
will produce a decreasing flux through loop. So current must be CCW.
IV.
A Capacitor C and a resistance R are connected in series across a battery as
shown. At the instant just after the switch is closed,
A) the voltage across C equals V.
B) the voltage across R is equal to V.
C) the voltage across R is zero.
D) the current is zero.
E) both C and D are correct.
Ans: B,
At t = 0, no charge is on the capacitor and so VC = 0 and all V
is across the resistor.
V.
In an electromagnetic wave in free space, the electric and magnetic fields are in
phase
A) True
B) False
Ans: A
VI.
A battery establishes a steady current around the circuit shown. What is the
relationship among the magnetic fields at points P,Q, and R?
A) BP > BQ > BR
B)
C)
D)
E)
BQ > BP > BR
BQ > BR > BP
BR > BP > BQ
BR = BQ > BP
Ans: B
B P is from one current, BQ is from two currents in same direction,
BR>is from two currents but in opposite directions.
VII.
A positively charged particle is moving northward in a magnetic field. The
magnetic force on the particle is northeast. What is the direction of the magnetic
field?
A) Up
D) South
B) Down
E) This situation can not exist.
C) West
Ans: E
Since F = qv x B, then F must always be perpendicular to B.
VIII.
Particle A of charge +2q is located at the origin, and particle B of charge +q is located on the
positive x-axis. The electric force on the particle B due to particle A is represented by vector F.
Which vector represents the force on particle A due to particle B?
A)
B)
C)
D)
F
-F
2F
-2F
E) 1/2 F
F) -1/2 F
G) The force is zero.
Ans : B , Newton's 3rd law
IX.
A solid insulating sphere, a solid conducting sphere and a hollow conducting
sphere are each given identical charge +Q. The spheres all have the same radius r.
At a distance a > r from the center of the spheres, the electric field
A) is greatest for the solid insulating sphere.
B) is greatest for the solid conducting sphere.
C) is greatest for the hollow conducting sphere.
D) is a non-zero value but the same for all three spheres.
E) is zero for all three spheres.
Ans: D
X.
A gaussian surface of radius a around each of the spheres will
enclose the same amount of charge +Q
James Clerk Maxwell made an important contribution to electromagnetic theory
by proposing that
A) an electric current is generated when a conducting coil rotates in a magnetic
field.
B) a time varying magnetic field generates an electric field.
C) magnetic monopoles are found in atomic nuclei.
D) magnetic fields are produced by time varying-electric field.
E) magnetism can be explained in terms of circulating currents within atoms.
Ans: D
1. A long straight wire, carrying a current I, is in the plane of a rectangular wire loop.
The loop has a length L; its sides parallel to the long wire are at distances a and b
from the wire as shown.
a) What is the magnetic field at a distance r (inside
the loop) from the long wire. Show the method and the
steps in this calculation.
b) What is the magnitude of the magnetic flux through
the loop.
c) Determine the maximum emf  induced in the loop by
the magnetic field created by the current in the loop, if I(t) = I0 sin(t), where I0
and  are constants.
d) Let a positive current I denote upward moving. What is the direction of the
induced current in the loop at t = 0?
Solution:
a) Using Ampere's Law: Choose an amperian loop of radius r around the
wire. From Ampere's Law  B  dl = 0 I
B is circular, and therefore at any point on loop, B and dl
Are tangent to circle and parallel.
I
r
Problem #1 (cont.)
 B dl cos 0  I
0
dl  rd
 B  dl  0 I
2
 B  rd  0 I
0

B
0 I
2 r
b) To find the magnetic flux through the loop:
I
 B   B  dA
where B dA and
dA  Ldr
0 I
 IL a b dr
Ldr  0 
a
2 r
2 a r
 IL
 IL a  b
a b
 B  0 ln r a
  B  0 ln
2
2
a
B  
c)
a b
dr
r
a
b
   ddt
B
where I (t )  I 0 sin  t
   2L ln a a b dtd ( I
0
   2L ln a a b ( I
0
 max

0
0
sin  t )
cos  t )
0 L I 0 a  b
ln
2
a
when cos  t  1
d)
    2L I
ab
cos  t
a
 I sin  t
BB 0 0
2 r
Although at t = 0 , B = 0, but B-field has the tendency to increase from
above (into the loop), and therefore increase flux through the loop. By
Lenz's law, the induced current will attempt to resist the positive change in
flux,. Therefore the induced current at t = 0 in the loop is CCW.
0
0
ln
2. A 30-pF air gap capacitor has circular plates of area A = 100 cm2 . It is charged by a
70-V battery through a 2.0- resistor. At the instant the battery is connected, the
electric field between the plates is changing most rapidly. At this instant, calculate (a)
capacitance, (b) time constant, (c) the current into the plates, (d) the rate of change of
electric field between the plates, (e) the magnetic field induced between the plates.
Assume E is uniform and is entirely contained between the plates, (i.e. E = 0 beyond
the edges of the plates).
Solution
a) C = 30 pF = 3.0  1011 F this value is just given
b)   RC    2.0 3.0 1011 F  6.0 1011 sec.
c)
The charge building on a capacitor when it is connected in series to a resistor is
given by:
Q  CV0 (1  et /  )
I
dQ
dt t 0
I
CV0


 I  CV0
CV0 V0

RC R
CV0 t / 
d
(1  e t /  ) 
e
dt

 at t  0  I 
t 0
70V

2.0
I  35 A
d) The electric field between the plates is:
E
Q
0 A
dE d Q
 (
) 
dt dt  0 A
From part c,
dE dE
1 dQ


dt
dt  0 A dt
dQ
 I  35 A
dt
so
dE
35 A

2
dt 8.85 1012 C
1.0 102 m2
N .m2
dE
V
 4.0 1014
dt
m.s
RC
e) B- lines are perpendicular to E-field and
E-out of page
r
is in circle around E.
B
From Maxwell-Ampere law:
d
 B  dl  0 I enc  0 0 dt E
The integration path is a circle of radius r < RC
Face of capacitor plate
Problem #2 (cont.)
B dl  and
2
dl  rd
B  rd  0 0 0
0
B
0 0 dE
2
r
dt
d
dE
( EA)  2 rB  0 0 ( r 2 )
dt
dt
where r  Rc
For r > RC the E = 0 beyond the edges of the plates. So  E   RC2 E ,( the flux is
entirely contained between the plates)
For r > RC , use the same procedure as above to obtain
d
2 rB  0 0 ( RC2 E )
dt
0 0 RC2 dE
2 rB 
2r dt
B
0 0 R 2 dE
C
2r
dt
where r  Rc
Bmax is when r = RC , from either of the above equations we obtain
B
0 0 R 2 dE
 Bmax 
C
2 Rc
dt
0 0 RC dE
2
 RC 
dt
A

and from part c,
dE
V
 4.0  1014
dt
m.s
1
T .m
C2
0.01m 2
V
Bmax  (4 107
 8.85 1012

 4.0 1014
2
2
A
N .m

m.s
Bmax  1.2 104 T
3. In the circuit shown, the switch is connected to position a for a long time, then thrown
to position b.
a) Find the frequency of oscillation in LC circuit.
b) Find the maximum charge on the capacitor.
0.1 H
c) Find the maximum current in the inductor.
d) Find the magnetic energy in the circuit at
1.0 F
S
b
3 ms after the switch is thrown to position b.
a
e) Find the total energy in the circuit 3 ms after
12 V
the switch is thrown to position b.
10 
Solution:
When Switch is at position b we have LC circuit.

1
a)  = 2f
where f 
and  
2
LC
1
f 
 f  503Hz
2 0.1H 1.0 106 F
b) After a long time that the switch has been closed to position a, the charge on the
capacitor will build up to maximum, when the potential across the capacitor will
equal the battery voltage VC  V .
Qmax  CV
 Qmax  12V 1.0 106 F

Qmax  12 F
c) When the charge on capacitor is zero, then the current in the inductor is
maximum, and this point the total energy of the circuit is stored in the inductor. On
the other hand when the capacitor is fully charged, then I = 0, and all the energy is
stored in the capacitor:
1 2
1
U L,max  U C ,max 
LI max  CVC2,max
2
2
I max  VC ,max
C
L
 I max  12.0V
1.0 106 F
0.1H
 I max  0.0379 A
d) The energy stored in the inductor after 3.0 ms is:
1 2
LI  I  I max sin  t
2
2
1
U L  (0.10 H ) 0.0379 A sin(2  503  3 103 sec) 
2
UL 
U L  2.67 107 J
Problem #3 (cont.)
e) The total energy of the circuit is constant for all times:
1
1 2
U total  CVC2  LI max
2
2
1
U total  0.10 H  (0.0379 A) 2   U Total  7.18 105 J
2
_________________________________________________________________
4. An insulating sphere of radius R 
2a
has a volume charge density  (r )  ar  br 2 ,
b
where a and b are constants.
a) What is the total charge contained in the volume of sphere?
b) What is the electric field at a distance of r from the center of the sphere.
c) What is the electric field at a distance of 2R from the center of the sphere.
d) At what radius does the electric field vanish?
Solution:
a) Given:
R
 (r ) 
dQ
dV
2a
b
and
 (r )  ar  br 2
 dQ   (r )dV
R
R
0
0
 where dV  4 r 2 dr
Q    (r )dV   (ar  br 2 )4 r 2 dr
R
R
 aR 4 bR 5 
Q  4   ar 3dr   br 4 dr   Q  4 

 0

0
5 
 4
Since R 
2a
b
 Q
16 a 5 128 a 5

b4
5b 4

Q
48 a 5
5b 4
b) Using a spherical shell of radius r for Gaussian surface:
Qenc
 E  dA 
0
r
1
 E  dA   
0
0
4 r 2 E 
E
1
0
r

r
0
 (r )(4 r 2 dr )
(ar  br 2 )(4 r 2 dr )
1  ar 2 br 3 

 0  4
5 
for
rR
Problem # 4 (cont.)
c) When r = 2R , the Gaussian surface of radius 2R will be chosen. In this case
all of the charge on sphere will be enclosed by gaussian surface. Using the
results of part (a)
 E  dA 
Qenc
0
4 (2 R) 2 E 
E
1
0
(
48 a 5
16a 2
1 48 a 5
)

4

(
)
E

(
)
5b 4
b2
0
5b 4
3a 3
4 0b 2
d) Using the result of part (b) to determine where the E-field vanishes:
E
1  ar 2 br 3 

0
 0  4
5 
ar 2 br 3

0 
4
5
r
5a
4b
5. A 400 turn long solenoid carries a current I (t )  30(1  e1.6t ) in A. While it is in
horizontal orientation, inside in the middle of this solenoid, and coaxial with it, a
smaller coil (solenoid) with 250 turns is placed. The long solenoid is 5.5 m long, and
has a radius of 16 cm, while the coil is 3 m long and has a radius of 6 cm.
a) What is the magnetic field inside the long solenoid?
b) What is the self-inductance of the long solenoid?
c) What is the magnetic flux through one of the turns of the smaller coil?
d) What is the mutual inductance?
e) What is the magnitude of the maximum emf induced in the coil?
Solution
The current on top of solenoid is out of page and on bottom into page.
a)
B
X X X X X X X X
Given:
N1=400, l1=5.5m, R1=0.16m,
N2=250, l2=3.0m,
R2=0.06m
Problem #5 (cont.)
a) The B-field inside a solenoid is given by
B   0 nI
B  0

where
n
N1
l1
N1
I 0 (1  e 1.6t )
l1
for t  1.0 sec

B  4  107
T .m 400
 30 A(1  e 1.6 )
A 5.5m
B  2.19  103 T
b) The self inductance for the long solenoid is found by:
N
L1  1 B
I
 B   B  dA  BA where B A
N1
IA
N1 BA
l1
L1 

I
I
2
0 N1 A
4 107 T .m / A  4002    (0.16m) 2
L1 
 L1 
l1
5.5m
N10
c) The flux through one tern of coil:

L1  2.94 103 H
 B ,2   B1  dA2  B1 A2
 B ,2 
0 N1 I 0 (1  e
1.6 t
where B1 A
2
) R
2
2
l1
for t  0 the B  field is found in part a   B ,2  2.19 103 T .(  0.062 m 2 )
 B ,2  2.48 105 T  m2
d) The mutual inductance is found by
N
M 12  2 B ,2
I
N 2 B ,2
M 12 
30 A(1  e1.6t )
at t  1.0sec  M 12 
250  2.48 105 T .m 2
30 A(1  e1.6 )

M 12  2.59 10 4 H
Problem #5 (cont.)
   M12 dI

dt
d
  M 12  30 A(1  e 1.6t ) 
dt
  M 12 (30 A 1.6e 1.6t )

 max is when t  0
 max  0.0124V


max
 1.6  30 A  2.59 104 H