Download PHY122 Midterm2

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Physics 122
Midterm Examination #2
March 22, 2006
Name:
Recitation Section:
Lab Section:
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Total
1. A conducting bar slides on rails separated by a distance l=1.5 m; the rails are
electrically connected with a resistance R=0.1 . There is a homogeneous
magnetic field B=1 T pointing into the ground (vertical into this page). The bar
moves with constant velocity v = 2 m/s to the left.
×
×
×
×
×
×
×
×
×
× ×
Fapp
×
×
×
×
×
×
×
×
×
×
a) Calculate the change of the magnet flux per
unit time for the loop connecting resistance
R and the sliding bar.
b) Find the induced potential difference (also
referred to as emf) across the bar.
c) What is the direction (clockwise or counter clockwise) and magnitude of the
current induced?
d) Find the magnitude of the constant external force pulling the bar Fapp.
Solution:
a) the change of magnetic flux with time is proportional to change of area (negative)
 B
m
Tm 2
  Bvl  1T 1.5m 2  3
and magnetic field,
t
s
s
b) According to Faraday’s law the induced emf is equal to to minus change of flux
emf  3V
c) Lenz’s law requires that induction opposes change, since change is due to
decreasing flux the induced magnetic field must point into plane and consequently
the current flows clockwise.
d) Force on a rode of length l caring a current I in magnetic field
emf
3V
is: F  I lB 
lB 
1.5m 1T  45 N
R
0.1
2. Consider the following AC circuit which is powered by a household outlet with
RMS voltage Vrms=120 V and frequency f=60 Hz. The circuit includes a resistor
with resistance R=200, an inductor with self inductance L=200mH, and a
capacitor with capacity C=20F.
a) Draw a phasor diagram for the circuit.
b) Calculate the impedance of the circuit.
c) Calculate the phase angle between
current and voltage in the circuit.
d) What is the power delivered by the AC source to this circuit?
e) What is the RMS voltage across the capacitor?
Solution:
a) need to calculate impedanze for Inductor and Capacitor:
X L   L  2 fL  2 60 Hz 0.2 H  75.4
XC 
1
1
1


 132.6
 C 2 fC 2 60 Hz 2  105 F
b) total impedance given by: Z  R2  ( X L  X C )2  208 
 X  XC
c) phase angle:   arctan  L
R


o
  17.7

d) need to consider phase angle in AC circuit:
V 2
Pav  I rmsVrms cos   rms cos   66W
Z
e) VCrms  X C I rms  76.5V
f) Two parallel wires carry current of 1 A and 2 A, respectively. The currents are
flowing in opposite direction. If the wires are separated by 10 cm find:
1.00 A
2.00 A
d=10cm
a) The location in horizontal
direction where the magnetic 2.00 A
field is zero.
b) The direction of the magnetic
field at P1 and P2.
c) The magnitude and direction of the force on a 10 cm long segment of the left wire
due to the magnetic field created by the right wire.
d) The magnitude and direction of the force on a 10 cm long segment of the right
wire due to the field created by the left wire.
Solution:
a) at x0 magnetic field will be zero if: B1 ( x0 )  B2 ( x0 )  0 . Here B1 is the
field at a distance x0 from wire with I1=2A and B2 is the field at a distance
x0+d from wire with I2=-1A.
0 I1
0 I 2
I2
I

0 
  1  I 2 x0   I 1 x0  I 1 d
2 x0 2 ( x0  d )
x0  d
x0
 x0  
I1
2A
d
10cm  20cm
I 2  I1
(  1 A)  2 A
B is zero 20 cm left of wire with 2A.
b) To get field at P1 and P2 contribution of both wires must be added
(use
P1:
0
 2 107
2
N
A2
):
B( x  10cm ) 
0  I 1
I 
N  2A
1 A 
 2   2107 2 


2  x x  d 
A  0.1m 0.2m 
B( x  10cm )  30  107 T
Right hand rule gives direction: out of plane
P2:
B( x  30cm ) 
0  I1
I 
N  2A
1 A 
 2   2107 2 


2  x x  d 
A  0.3m 0.2m 
B( x  30cm )  3.3  107 T
Right hand rule give direction in to plane
c) Calculate force on 10 cm segments of wire with I2=1A due to magnetic
field around wire with I1=2A. First calculate field:
B(d )  
0 I1
2 d
force on wire segment now is:
F  I 2 lB(d ) 
0
l
N
10cm
I1 I 2
 2  107 2 (1 A)2 A
 4  107 N
2
d
A
10cm
according to right hand rule force points to left.
d) Opposite direction and equal magnitude.
2. A positive charge of 1 C and a mass of 1 g is accelerated through a potential
difference V = 1000V and shot into a magnetic field B=1T pointing vertically into
the plane.
1km
V
Solution c)
a) What is the velocity of the charge after it leaves the plates but before it reaches
the magnetic field?
b) Calculate the radius of curvature of the path the charge will take in the magnetic
field.
c) Draw the path approximately to scale.
Solution:
a) Charge accelerated through electrical potential V, convert potential to kinetic
energy:
PE  qV  106 C 1000V  103 J
1
PE  KE  mv 2
2
v
2 PE

m
2  103 Nm
m
 1.41
3
10 kg
s
b) Force due to magnetic field provides centripetal force:
mv 2
mv
FB  qvB 
 R
 1410 m
R
qB
The primary side of a transformer is powered by an AC source. The coil on the
primary side has N1=4 loops while the one on the secondary side has N2=8 loops.
AC power source
Vrms = 20 V
f = 50 Hz
R = 100
a) What is the maximum (or peak) voltage and frequency on the secondary side?
b) Calculate the power delivered to the resistor on the secondary side.
c) Find the current flowing through the loop on the primary side.
Solution:
a) Frequency unchanged, i.e. 50 Hz. Peak voltage on secondary
N
side: V2max  2V2 rms  2 2 V1rms  56.6V
N1
b) Pav 
V2 rms 2
 16W
R
c) power on primary and secondary side the same:
I1rms 
Pav
 0.8 A
V1rms