Download Exam 2 Solutions

Name:_______________________ ___
PHY2061
11-8-05
Exam 2 Solutions
1. An RC circuit is discharged by closing a switch at time t = 0. The initial potential
difference across the capacitor is 5 V. The potential difference across the capacitor drops to
half of its value in 35 ms.
(a) [6 points] What is the time constant of the circuit?
The expression for the potential difference across the capacitor when discharging is
given by VC   e t / where   RC is the time constant
If the voltage drops to half of its value,

  e  t /
2
2  et / 
ln 2  t / 

t
35 ms

 50.5 ms
ln 2 0.693
(b) [4 points] If the total resistance in the circuit is 10 k, what is the capacitance?

50.5 103 s
  RC  C  
 5.05 F
R
10000
Page 1 of 11
Name:_______________________ ___
PHY2061
11-8-05
2. [8 points] In the circuit shown, the resistances are R1  6 , R2  12 , and R3  24 . The
battery voltages are 1  18 V and  2  6 V . What is the current (in amps) flowing through
the battery with potential difference 1 ?
i2
i1
i3
Let i1 be the current through the 1, 18V, battery, and i2 the current through the right branch
with the 2, 6V, battery, and i3 the current through the middle branch.
By Kirchoff’s junction rule: i1  i2  i3 .
By Kirchoff’s voltage sum rule applied to the rightmost loop:
 2  i2 R2  i3 R3  0 (6V  12i2  24i3  0)
i3 
R2

i2  2
R3
R3
(i3  0.5i2  0.25)
By Kirchoff’s voltage sum rule applied to the outer loop:
1  i1 R1  i2 R2   2  0 (18V  6i1  6V  12i2 =0)
1  R1  i2  i3   R2i2   2
1  i2 R1 

1   1 
R1 R2
R
i2  1  2  R2i2   2
R3
R3

R1 
R1 R2 
  2   R1  R2 
 i2
R3 
R3 


10.5
i2 
 0.5
21
12
6
i3  i2 
 0.5
24
24
 i1  i2  i3  1.0
Answer: 1.0 Amps through 18V battery
Page 2 of 11
Name:_______________________ ___
PHY2061
11-8-05
3. [6 points] Two identical amounts (volumes) of copper are formed into two cylindrical
rods, the first of length L and the second of length L/2. If the resistances of the two volumes
(measured between the two ends) are R1 and R2 respectively, what is R2 / R1 ?
Resistance is given by the expression:
L
L2
R 
where V  LA  volume and   resistivity of copper
A
V
The key point is that the volume remains the same, so the cross sectional area must change.
Thus
2
R2   L2 / V  L2 / 4 1

 2 
R1   L12 / V 
L
4
4. [6 points] A large flat non-conducting sheet has an electric charge density of
  5.0C / m2 , where 1C  106 C , as measured in the rest frame of the sheet. What is the
charge density “seen” by an electron traveling at a velocity of v  2.9 108 m/s parallel to the
surface of the sheet?
Length is contracted along the direction of motion: L 
L0

, but not in directions
perpendicular to the direction of motion. However, the electric charge in some region (let’s
say a square in the rest frame of the sheet), must remain invariant between different
reference frames. So if the square area on the sheet gets contracted along the direction of
motion (becoming a rectangle as observed by the electron), the area decreases while the
charge remains the same. So the charge density should increase.
In the rest frame of the sheet, the area of the square is L20 and the charge is q   0 L20 , where
0 is the charge density in the rest frame.
In the frame of the electron, the sheet area is contracted to: A 
L0 2

The observed charge density is thus:
q  L2
   20 0   0
A L0 / 
That is, the charge density appears higher to the electron by the factor . Plugging in:
1

 3.9
v2
1 2
c
   3.9  5 C / m 2   19.5 C / m 2
Page 3 of 11
Name:_______________________ ___
PHY2061
11-8-05
5. [6 points] Galaxy A is reported to be receding from us at a speed of 0.41c. Galaxy B,
located in precisely the opposite direction, is found to be receding from us at the same speed.
What recessional speed would an observer on Galaxy A find for Galaxy B?
The velocity of Galaxy B in the rest frame of us on Earth is 0.41c. We want to transform this
velocity to a frame moving at a velocity of -0.41c moving in the opposite direction.
Use the Lorentz transformation of velocity:
u   v 
u x  x
1  ux v / c 2
u x 
0.41c  0.41c
 0.702c  2.11108 m/s
1   0.41 0.41
Note that this is not the Galilean expectation of 0.82c.
6. [6 points] An electron is accelerated from rest by a potential difference of 10,000 V. It
then enters a uniform magnetic field of magnitude 0.05 T with its velocity perpendicular to
the direction of the field. Calculate the radius in meters of its path in the magnetic field. The
mass of the electron is 9.111031 kg .
Setting the centripetal acceleration, a  v 2 / r , equal to the acceleration due to the magnetic
force, FB / m  qvB / m , yields the following relation for the radius of curvature for a charged
particle moving in a uniform magnetic field:
mv
p
r

qB qB
To get the velocity, one must realize that the kinetic energy of the electron after accelerating
from rest across an electric potential equals the initial potential energy:
1
EK  mv 2  eV
2
 mv  2meV
2  9.1110 31 kg 1.6 10 19  10000 V 
2meV
r

 6.75 10 3 m
19
qB
1.6 10   0.05
Answer: 6.75 103 m
Page 4 of 11
PHY2061
11-8-05
Name:_______________________ ___
7. The rectangular-shaped loop shown in the figure is carrying a current i = 2A in the
clockwise direction shown. The length of the vertical sides is 7cm and the length of the
horizontal sides is 3.5 cm. A uniform magnitude field B = 0.05T makes an angle =30 with
respect to the plane of the plane of the loop.
(a) [6 points] What is the magnitude of the torque acting on the loop
if the left side is connected to a hinge?


Note, by the right-hand rule, the magnetic dipole moment μ  iA points in the zˆ direction.
The torque is given by:
τ  μB

   B sin 
where    
2
   2 A  0.035 m  0.07 m  0.05 T  cos  30   2.1 10 4 N-m
(b) [6 points] If the loop is free to rotate about its hinge due to the torque, by how much will
the potential energy of the loop change when the torque becomes zero?
Potential energy of a dipole in a magnetic field is given by:
U  μ  B
U min    B




U    U        B cos        B ( sin  )
2

2

U  U    U min   B  sin   1  1.5iAB  3.7  10 4 J
Page 5 of 11
PHY2061
11-8-05
Name:_______________________ ___
8. [8 points] A long straight wire carries a current i1  80A in the horizontal direction shown.
Below it is a square loop of side length b  50 cm carrying a current i2  40A in the
clockwise direction shown. The distance between the top of loop to the long straight wire is
distance a, which you can take to be much smaller than b ( a b ). If the mass of the loop is
m  10 g, what is the maximum distance a such that the loop is attracted to the wire with
current i1 with a force that overcomes the force of gravity? (the acceleration due to gravity
points in the yˆ direction).
b
b
The magnitude of the force between two parallel current-carrying wires is:
F
0 Li1i2
2 d
The force on the top wire is attractive (up) since the currents are in the same direction, and
the force on the bottom wire is repulsive (down). By symmetry, the force on the two vertical
sides will cancel (and are actually zero because L  B  0 ).
The net force is up since the distance to the bottom wire is larger:
0 Li1i2  1
1  0 Li1i2
 mg
 

2  a a  b 
2 a
 Li i
 a  0 1 2  0.00327  3.3 mm
2 mg
F 
Page 6 of 11
Name:_______________________ ___
PHY2061
11-8-05
9. [6 points] A horizontal power line carries a current of 3000A from south to north. Earth's
magnetic field, with a magnitude of 60T, is directed toward the north with a dip angle 60°
downward into the Earth relative to the horizontal. Find the magnitude and direction (use
compass directions) of the magnetic force acting on a 100 m length of power line.
The magnetic field and the power line both point north, but the magnetic field points into the
Earth at an angle of 60°.
i
S
60°
B
The force on a current carrying wire is given by:
F  iL×B
So the force points west with magnitude:
F  iLB sin    3000 100   60 106   sin 60   15.6 N
Answer: 15.6 N
Page 7 of 11
N
Name:_______________________ ___
PHY2061
11-8-05
10. [8 points] An infinitely long insulated wire carrying a current I = 100 A is bent into the
shape shown (straight line plus circle of radius R=3 cm with the currents in the direction
shown). Find the magnitude of the field B at the center of the circle.
We calculate the field from both the long wire and the circle separately. From the long wire,
the field at the center of the circle, a distance R away, is:
i
Bwire  0  6.7 104 T into the plane of the paper.
2 R
From the circular wire, the field is:
i
Barc  0  where   2
4 R
i
Barc  0  2.1103 T out of the plane of the paper
2R
Since the magnetic field from the arc is in the opposite direction of that from the long wire,
the total magnitude is the difference:
B  Barc  Bwire 
0i   1 4 107 100   1

 1.4 103 T
2 R
2  0.03
Answer: 1.4 103
Page 8 of 11
Name:_______________________ ___
PHY2061
11-8-05
11. [6 points] A conductor has a cylindrical cross section with a radius a = 2 mm.
Inside, a cylindrical cut-out hole (non-conducting) is made as shown, where the radius of
the cut-out is b = 0.5 mm and the distance between the center of the main conductor and
the center of the cut-out is d = 1 mm. The magnitude of the current density in the
conductor is j  105 A/m2 . What is the magnitude of the magnetic field at a radius of
r  0.5 mm from the center of the conductor along a line connecting to the center of the
cut-out? [Hint: try using the superposition principle.]
We break this problem up into two contributions: the magnetic field due to a solid conductor
of radius a and with a current coming out of the page, and the magnetic field from a smaller
conductor of radius b with a current going into the page. (The two currents will cancel,
leading to an effective cavity in the shown conductor).
By Ampere’s Law, the field inside the large cylindrical conductor at a radius r can be
obtained from:

a
B  ds  0 ienc
2 rBa  0 j r 2
Ba 
0 j
r  3.14 105 T
2
It points in the counter-clockwise direction for a current out of the page and for r=0.5 mm to
the right of the center.
The field from the smaller conductor has the exact same magnitude since the current density
is equal and opposite. Also, the radius r=0.5mm places the point for the field to be calculated
at the same radius from the center of conductor b. Thus:
Bb 
0 j
2
r
It points in the clockwise direction, but at r=0.5mm from the center of conductor a it points
in the same direction! So the two fields add, and we have:
B  0 jr  4 107 105   0.0005   6.3 105 T
Page 9 of 11
Name:_______________________ ___
PHY2061
11-8-05
12. A rectangular loop of wire of sides L  2 L is moving with constant speed v into a region
of uniform magnetic field of magnitude B directed out of the paper as shown in the figure
(the long side is along the direction of motion).
(a) [6 points] What is the magnitude of the induced EMF in the
wire and the direction of the induced current just
after the loop enters the region of magnetic field?
L
dB
dt
The magnetic field is a constant, but the area in the field region changes:
Faraday’s Law:   
A  L  c  vt  , so
 
dB
d
 BL  c  vt   BLv
dt
dt
By Lenz’s Law, the current is in the direction such that it creates a magnetic field that
opposes the changing flux. Since the flux is increasing through the loop out of the
page, we need to add a field into the page which can be accomplished with a
clockwise current.
(b) [6 points] If the total resistance of the wire is R, what is the power dissipated by the
loop?

BLv
R
R
B 2 L2 v 2
P  i2 R 
R
i

Page 10 of 11
Name:_______________________ ___
PHY2061
11-8-05
13. [6 points] A circular loop of wire of radius 30 cm 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. The magnetic field is now
increased at a constant rate such that it reaches three times its original strength in 20 s. What
is the magnitude of the induced EMF in the wire loop during this time?
dB
dt
The loop area remains the same, only the magnetic field changes:
Faraday’s Law:   
 3B0  B0 
dB
dB
B
A
  r2
  r2
dt
dt
t
t
2 2  0.5 T 
    0.3 m 
 0.014 V
20s
 
Page 11 of 11