Scalar Product:
A  B  A B cos  AB  Ax Bx  Ay By  Az Bz
Vector Product:
A  B  A B sin  AB



  r F
Newton’s second law for rotation



 I
Conditions for rolling:
  Ay Bz  Az B y  iˆ
aCOM   R and vCOM  R
  Az Bx  Ax Bz  ˆj
Angular momentum:
  Ax B y  Ay Bx  kˆ
Equations of motion:
v  v0  at
1
x  x0  v0t  at 2
2
2
2
v  v0  2a( x  x0 )
Radial Acceleration:
arad 
v2
r
L  I , where I   mi ri2
i
Newton’s Law of Gravitation:
Gm1m2
Gm1m2
and U G  
with
2
r
r
UG  0 at infinity
FG 
Bernoulli's Equation:
1
1
p1   gy1   v12  p2   gy2   v22
2
2
Equation for Simple Harmonic Motion:
Newton’s second law

L  r  p or

 F  ma
Magnitude of kinetic friction
F f k   k FN
Magnitude of static friction
d 2x
  2 x
2
dt
Solution for above equation:
x  t   A cos t   
Where,
2
T
Ffs  s FN
  2 f 
Definition of work
For a spring mass oscillator,
Definition of kinetic energy:

 
W   F  dx
KE 
1 2
mv
2
For a simple pendulum,
Change in gravitational potential energy:
U g  mg y
Elastic potential energy:
U el 
1 2
kx
2
Work-Energy Theorem:
W  U  KE
Center-of-mass position
X COM 
1
M
k
m
n
x m
i 1
i
i
Definition of momentum


p  mv
Conservation of momentum


pi  p f
Definition of torque

g
L
Wave Equation:
 2 y  x, t  1  2 y  x, t 
 2
x 2
v
t 2
Solution to above equation:
y  x, t   A cos  kx  t 
Where,
2
k
,   2 f , v   f

Standing waves on fixed string:
y  x, t   ASW sin  kx  sin t 
fn  n
v
2L
Doppler Effect:
1 qQ
rˆ
4 0 r 2
q
 E   E  dA  enc
FE  qE 
v  vL
fL 
fS
v  vS
5
T oC  T o F  32
9
T  K   T oC  273.15
 
  

0
 
b
Vb  Va    E  dl
L   L0 T
a
V  V0 T
Q  mcT  nC T
QF V   mLF V
dQ
A
 k TH  TC 
dt
L
pV  nRT
3
Ktr  nRT
2
3
CV  R ideal monatomic gas
2
5
CV  R ideal diatomic gas w/o vibration
2
H
W
V2
 pdV
V1
U  Q  W
U  nCV T
for ideal gas
pV   const 
 adiabatic process
TV  1  const 
e
W
Q
 1 C
QH
QH
eCarnot  1 
TC
TH
U  qV
 V ˆ V ˆ V 
E    iˆ
j
k

y
z 
 x
Q  CV
1
1
1
1
 


series
Ceq C1 C2 C3
Ceq  C1  C2  C3 
parallel
1
U  CV 2
2
1
uE   0 E 2
2
E0
E
K
dq
I
dt
J  nqvd
E

J
L
R
A
V  IR
P  VI
Req  R1  R2  R3 
series
1
1
1
1
 
 
Req R1 R2 R3
2
dQ
S  
T
1
S  k ln w
R  8.314 J mol  K
qC
q  Q0e
N A  6.02 10 molecules mole
23
2
5
1 atm = 101 325 N / (m ) = 1.01x10 Pa
1/4πε0 = 8.99 x 109 Nm2/C2
e = -1.602 x 10-19 C
1 e 
t
t
RC
RC
parallel
charging
discharging
F  qv  B
 B   B  dA
dF  Idl  B
    B ,   NIA
U    B
 qv  rˆ
B 0
4 r 2
 Idl  rˆ
dB  0
4 r 2
 B  dl  0 Ienc
Vs  Vp
 B  dl   i
 iD 
   dl 
dB
dt
0
C

dE
dt
di1
di
and 1   M 2
2  M
dt
dt
N 2  B 2 N1 B1
M

i1
i2
di
 L ,
dt
NB
L
i
1 2
1
uE 
B
U  LI 2
2

2
0
,
di
 Rt
 e L
dt L
1

LC
1
I RMS 
I for i  I cos t 
2
1
VRMS 
V for v  V cos t 
2
VR  IR
iD  
VL  IX L , where X L   L
VC  IX C , where X C 
1
C
V  IZ , where Z  R 2   X L  X C 
X  XC
1
PAvg  VI cos  , tan   L
2
R
2
Ns
Np
Physics 161-001 Spring 2014 Exam 3
Name: __________________________________ Box#_____________
Multiple Choice (5 points each):
1) A tube of mercury with resistivity 7.84 × 10-6 Ωm has an electric field inside the
column of mercury of magnitude 8 V/m that is directed along the length of the
tube. How much current is flowing through this tube if its radius is 6.0 mm?
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
4.80 A
6.00 A
10.0 A
12.3 A
19.2 A
25.0 A
55.4 A
87.3 A
115 A
134 A
J
E


8V m
 1.0  106 A m2 and
7.84  106 m


I  J  A  1.0  106 A m2    6  103 m   115 A .
2
2) The emf and the internal resistance of a battery are as shown in the figure. When
the terminal voltage Vab is equal to 21.0 V, what is the current through the
battery?
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
1.2 A
4.3 A
5.0 A
10.0 A
15.5 A
19.1 A
23.2 A
26.0 A
52.0 A
104 A
Vab  21.0V  E  Ir  31V  I  2  
I  5.0 A
3) A proton moving in the positive x direction enters a magnetic field. If the proton
experiences a magnetic deflection in the negative y direction, the magnetic field
in this region is
A)
B)
C)
D)
E)
F)
G)
H)
I)
in the direction of the +x axis.
in the direction of the -x axis.
FB  qv  B . Since the proton is positively charged,
in the direction of the +y axis.
the direction is in the same direction given by the right
in the direction of the -y axis.
hand rule.
in the direction of the +z axis.
in the direction of the -z axis.
in any direction perpendicular to the proton velocity.
zero.
undefined.
4) If the current density in a wire of radius R is given by J = J0 + kr, 0 < r < R, what
is the total current in the wire?
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
kR2/2
kR2
J0R + kR2/2
J0R2 + kR3/3
J0πR2 + k2πR3/3
J02πR + k2πR2/2
J0πR + k2R2/2
k2πR3/3
kR3/3
J0(k2πR3/3)
R
R
0
0
I   J  r  dA   J  r  2 rdr
R
R
R
0
0
   J 0  kr  2 rdr   J 0 2 rdr   kr 2 rdr
0
R
R
 J 0 2  rdr  2 k  r 2dr
0
0
 J 0 R  2 k R 3
2
3
5) The figure shows the cross-section of a hollow cylinder of inner radius a = 1.0 cm
and outer radius b = 2.0 cm. A uniform current density of 1.0 A/ cm2 flows
through the cylinder parallel to its axis. Calculate the magnitude of the magnetic
field at a distance of d = 1.0 m from the axis of the cylinder. (μ 0 = 4π × 10-7 T ·
m/A)
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
1.1 x 10-5 T
1.5 x 10-5 T
1.1 x 10-6 T
1.9 x 10-6 T
6.3 x 10-6 T
9.7 x 10-6 T
2.2 x 10-7 T
4.8 x 10-7 T
9.1 x 10-7 T
0T
In this case, J is uniform, so,
I  JA  J  b2   a 2   1 A cm 2    b2  a 2 
 1 A cm2    4cm2  1cm2   9.4 A
Then, from symmetry, we use Ampere’s law to find
that:


 B d s   i
0 enc

B 2 r  4  107 T  m
A
  9.4 A 
B  1.9  106 T
6) Calculate the current through a 1.0-m long 22 gauge (having radius 0.321 mm)
nichrome wire if it is connected to a 3.0-V battery. The resistivity of nichrome is
100 × 10-8 Ω · m.
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
1A
2A
3A
4A
5A
6A
7A
8A
9A
10 A
R
L
A

100  108 m  1m
  3.21  104 m 
and,
V  IR 
I
V
3V

 1A
R 3.1
2
 3.1
7) The figure shows three identical light bulbs connected to a battery having a
constant voltage across its terminals. What happens to the brightness of light bulb
3 when the switch S is closed?
A)
B)
C)
D)
E)
Momentarily goes up then back to its original brightness.
Momentarily goes down then back to its original brightness.
Permanently gets brighter.
Permanently gets dimmer.
No change.
8) A current is running through a wire next to the circuit shown in the figure with the
switch S open and the capacitor uncharged. The battery has no appreciable
internal resistance. Which one of the following graphs best describes the
magnitude of the force on the wire as a function of time t after closing the switch?
F
F
B
A
F
F
C
F
E
D
9) For the circuit shown in the figure, determine the current in the 4.0-Ω resistor.
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
0.1 A
0.2 A
0.4 A
0.6 A
0.8 A
1.0 A
1.2 A
1.4 A
1.6 A
1.8 A
2.0
2.0
1.2
We first find the equivalent resistance of the three resistors in parallel:
1
1
1
1
4



 Req   , then the equivalent resistance of
Req 2 2 2
5
4
the three in series: Req    2  1.2  4
5
and, then the current through the entire circuit: I = V/R = 3A. Then, the
voltage across the resistors in parallel is 12/5V and so the current
through the 4.0-Ω resistor is 3/5 A.
10) Consider the circuit shown in the figure. Note that two currents are shown.
 IR 
Calculate the emfV ε3.
V
3V
I 
 1A
R 3.1 2.0 A
10.0
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
50 V
48 V
44 V
42 V
40 V
38 V
36 V
34 V
32 V
30 V
Vab  21.0V  E  Ir  31V  I  2  
I  5.0 A
11) For the circuit shown in the figure, the switch S is initially open and the capacitor
voltage is 80 V. The switch is then closed at time t = 0. What is the charge on the
capacitor when the current in the circuit is 13 μA?
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
110 μC
140 μC
200 μC
220 μC
280 μC
330 μC
390 μC
430 μC
470 μC
500 μC
q  Q0e t RC and i 
dq
Q
  0 e  t RC .
dt
RC
Now, at some time, i = 13μA, or:
Q0  t RC
e
 13 A 
RC
RC
RC
R
e  t RC 
13 A 
13 A  13 A
Q0
CV0
V0
i
then,
q  Q0e  t RC  Q0
R
13 A  C  R  13 A  430C
V0
12) For the circuit shown in the figure, the capacitors are all initially uncharged, the
connecting leads have no resistance, the battery has no appreciable internal
resistance, and the switch S is originally open. Just after closing the switch S,
what is the current in the 15.0-Ω resistor?
30.0
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
0A
0.2 A
0.4 A
0.6 A
0.8 A
1.0 A
1.2 A
1.4A
1.6 A
1.8 A
Immediately after the switch is closed, the voltage
drop across the capacitor is 0V, so no current
through the 15 Ω resistor.
13) After the switch S has been closed for a very long time, what is the potential
difference across the 28.0-μF capacitor?
30.0
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
0.0 V
4.3 V
5.0 V
10.0 V
15.0 V
19.1 V
23.2 V
26.0 V
30.0 V
104 V
For long times, no current passes through the
capacitors, so the current just goes through the three
resistors in series. The equivalent resistance is just
30 Ω, with 30V means 1 A is passing through each
resistor. Then the voltage drop across the 15 Ω
resistor is just 15 V.
14) A charge is accelerated from rest through a potential difference V and then enters
a uniform magnetic field oriented perpendicular to its path. The field deflects the
particle into a circular arc of radius R. If the accelerating potential is doubled to
2V, what will be the radius of the circular arc?
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
2R
R/2
4R
R/4
2R
R/ 2
R
3R
3R
R/ 3
1 2
mv  qV 
2
and
2qv
v
m
FB  qv  B 
KE 
v2

R
mv 2 mv m 2qV
1
R



qvB qB qB
m
B
 qvB  ma  m
2mV
q
15) A circular coil of wire of 100 turns and diameter 10.0 cm carries a current of 1.0
A. It is placed in a magnetic field of 1 T with the plane of the coil making an
angle of 45° with the magnetic field. What is the magnetic torque on the coil?
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
2.0 Nm
2.2 Nm
2.4 Nm
2.6 Nm
2.8 Nm
3.0 Nm
3.2 Nm
3.4 Nm
3.6 Nm
3.8 Nm
    B and
  NIA  100  1.0 A    0.1m   3.14 Am2 so
2
   B sin   3.14 Am2  1T  sin 45  2.22 Nm
16) A negatively charged particle is moving to the right, directly above a wire having
a current flowing to the right, as shown in the figure. In which direction is the
magnetic force exerted on the particle?
A)
B)
C)
D)
E)
up
down
out of the page
into the page
no magnetic force
The B-field from the wire is coming out of the page
at the charge, and from F  qv  B , with q
negative, then the force is upwards.
17) A large number of very long wires of diameter 1mm are laid side-by-side to form
a plane. If 10.5 A of current is passed through each wire (in the same direction),
what is the magnitude of the magnetic field 10cm above (and in the middle of) the
plane? (μ 0 = 4π × 10-7 T · m/A)
A)
B)
C)
D)
E)
F)
G)
H)
I)
J)
1.6 x 10-4 T
5.2 x 10-4 T
7.9 x 10-4 T
2.6 x 10-3 T
4.7 x 10-3 T
6.6 x 10-3 T
9.2 x 10-3 T
2.1 x 10-2 T
4.4 x 10-2 T
6.5 x 10-2 T


 B d s  0ienc

2 BL  4  107 T  m

B  2  107 T  m
 6.6  103 T
10.5 A L 


A 1  10 m
3
10.5 A


A 1  10 m
3