Physics 260
Exam 2
TR 12:45 – 2:05
Dr. Womble
Name:_______________________________
THE MORE WORK SHOWN, THE MORE CREDIT GIVEN!
Constants:
1
k=
= 9 x 109 Nm2/C2
4o
o = 8.85 x 10-12 C2/ Nm2
0 = 4 x 10-7 Tm/A
mass electron, me = 9.11 x 10-31 kg
electron charge, e = -1.6 x 10-19 C
Conversions:
1 eV =1.6 x 10-19 J
1 u = 1.66 x 10-27 kg
1) A thin metal bar with a mass of 10 g and length of 5.0 cm rests on, but is not attached
to, two metallic supports in a uniform magnetic field with a magnitude of 1 T, as
shown in the figure below . A battery and a resistor of resistance 50.0  are
connected in series to the supports. What is largest battery that can be used without
breaking the circuit? (20 pts)
F=iLB
F=mg
V=iR
Max force is when mg=iLB
But i=V/R so
mg=V*L*B/R
V=mgR/(L*B)
V= 0.01*9.8*50/(0.05*1)
2) Maxwell’s Equations:
a) (5 pts) Discuss the physical meaning of the following equation:
  q
 
Integral Form:  E  dA  enclosed
Differential Form:   E  enclosed
0
0
b) (5 pts) Discuss the physical meaning of the following equation:
 

Integral Form:  B  dA  0
Differential Form:   B  0
c) (5 pts) Discuss the physical meaning of the following equation:
 
 
d
Integral Form:  E  ds    B  dA
dt


dB
Differential Form:   E  
dt
d) (5 pts) Discuss the physical meaning of the following equation:
 
 
d

Integral Form:  B  ds   0  ienclosed   0  E  dA 
dt




dE 

Differential Form:   B  0  J enclosed   0
dt 

3) A rectangular coil of 10 turns and of length 10 cm and width, 25 cm, is rotated at
frequency of 60 Hz in a uniform magnetic field B= 10 mT as shown in the figure
below. The coil is connected to co-rotating cylinders against which metal brushes
slide to make contact. If the co-rotating cylinders are attached to a 50  resistor,
what is the current flowing through the resistor (Hint: as a function to t)? (20 pts)
B
50 
PHI=N*B*A*cos(theta); theta=omega*t and omega=2*pi*f
N= 10 turns, B=0.01 T, A=L*W=0.1*0.25=0.025 m2, f=60 Hz
EMF=-d/dt(PHI)= -N*B*A* d/dt(cos(omega*t))
d/dt(cos(omega*t)=omega*(-sin(omega*t))
EMF=N*B*A*omega*sin(omega*t)
EMF=i*R
i = N*B*A*omega*(1/R)*sin(omega*t)
4) A circular loop of wire with a radius of 11.4 cm and oriented in the horizontal xyplane is located in a region of uniform magnetic field. A magnetic field with a
magnitude of 1.20 T is directed along the positive z-direction, which is upward.
a) If the loop is removed from the field region in a time interval of 2.40×10−3 s, find
the average emf that will be induced in the wire loop during the extraction process
(15 pts)
DeltaT=2.4e-3 Phi_initial=B*A where B=1.2 T, and A=pi*r^2 where r=.114 m
DeltaPhi=Phi_final-Phi_initial, Phi_final=0
DeltaPhi=-B*A
EMF=-DeltaPhi/DeltaT=B*A/DeltaT
b) If the coil is viewed looking down on it from above, is the induced current in the
loop clockwise or counterclockwise? (5 pts)
DeltaPhi<0 so EMF>0 therefore CCW
5) The figure below shows a cross-section of a coaxial cable and
gives its radii (a,b). The current in the inner conductor is -3i
but the current in the outer conductor is +2i. However, the
current in the inner conductor is uniformly distributed (i.e. a
constant current density). The outer conductor may be treated as
thin shell. Derive expressions for the magnitude of the magnetic
field B(r) for the following regions in:
Hint: The left hand side of Ampere’s Law for this geometry is
 
B
  d s  B(2r )
a) r < a (10 pts)
J=-3i/(pi*a^2) i_enc=J*A where A=pi*r^2
B(2*pi*r)=mu_0*(-3i/(pi*a^2)*pi*r^2
B=-3*mu_0*r/(2*pi*a^2)
b) a < r < b (7 pts)
B=-3*mu_0*i/(2*pi*r)
c) r > b (3 pts)
B=-mu_0*i/(2*pi*r)
a
b
List of Necessary and Unnecessary Equations
1.
F=
1
q1 q 2
4 o
r2
19. P= iV = i2R =
mv 2
20. Fc =
r
2. F=qE
1
q
r̂
4o r 2
F
E=
q
3. E =
4.
5.  E 
21. K = ½ mv2
22. L = mvr
23. T =
 E  da
24. f=
 o   qenc

7. E=
o

8. E=
2 o

9. E =
2o r
 oi
2r
 i
30. B = o
4r
31. B= 0 i n

i
f
E  ds
34.
13.
V
i 1
i

1
4o
V
14. Es= s
15. i =

J  dA
V
i
1 E
17.   
 J
L
18. R = 
A
16. R=
32.
 B  ds   i
0 enc
33. Fba= ibLBa
q
12. V=
4 o r
V=
 o ids  rˆ
4 r 2
29. B =
1
n
1
T
28. dB =
W
q
11. Vf - Vi = 
2r
v
25. F = q v  B
26. F = i L  B
 = N i A
6.
10. V = 
V2
R
n
35.
qi
r
i 1
i
 B  dA  0
d
 E  ds  - dt
B
36.
 B  ds  
0
 0 dE   0 i enc
dt
37. E= - dB
dt
38. id =  0 dE
dt
39. E = i R
40. E = Blv
41.  B 
 B  da
42. =2**f