Recitation Week 7
Chapters 26 and 27
Problem 26.61. Calculate the three currents I1 , I2 , and I3 indicated in the circuit diagram shown in Fig. 26.65.
5.00
8.00
+------/\/\/------+------/\/\/------+
|12.00V 1.00
I2 | I1 1.00 9.00V |
+--|i---/\/\/--<--+-->--/\/\/---i|--+
|
I3
10.00
|
+-------->------/\/\/---------------+
Start by labeling everything.
5.00=R5 I5
I4 8.00=R4
+------/\/\/--->--+--<---/\/\/------+
|12.0V 1.00=R2 I2 a I1 1.00=R1 9.00V|
b--|i---/\/\/--<--+-->--/\/\/---i|--c
|
I3
10.00=R3
|
+-------->------/\/\/---------------+
Then junction rules at a and b yield
0 = −I1 − I2 + I5 + I4
(1)
0 = I2 − I3 − I5
(2)
Loop rules for each sub-loop clockwise from the loop’s upper left yield
0 = −I5 R5 − I2 R2 + V2
(3)
0 = +I4 R4 − V1 + I1 R1
(4)
0 = −V2 + I2 R2 − I1 R1 + V1 + I3 R3
(5)
Which gives us five equations for the five unknown currents. Solve using your method of choice (e.g. linear algebra).

 
 
I1
0
−1
−1
0
1
1
 0   0
 I2 
1
−1
0
−1

 
 
 
 −V2  =  0
−R2 0
0 −R5 

 I3 
 
 V1   R1
0
0 R4
0  I4 
−R1 R2 R3 0
0
I5
V2 − V1
  
−1 
 

I1
−1
−1
0
1
1
0
0.848
I2   0

 

1
−1 0
−1 
  
  0   2.14 
I3  =  0




−R2 0
0 −R5   −V2  = 0.171
  
 A
I4   R1
0
0 R4
0   V1   1.02 
I5
−R1 R2 R3 0
0
V2 − V1
1.97
(6)
(7)
Problem 26.86. An R-C circuit has a time constant RC. (a) If the circuit is discharging, how long will it take for its stored
energy to be reduced to 1/e of its initial value? (b) If it is charging, how long will it take for the stored energy to reach 1/e
of its maximum value?
The energy stored in the capacitor is 1/e of its maximum value when
U=
This value is useful for both (a) and (b).
U0
Q2
q2
=
= 0
2C
e
2Ce
2
Q
q2 = 0
e
q = Q0 e−0.5 .
(8)
(9)
(10)
(a) For a discharging circuit,
−t
q = Q0 e RC ,
(11)
so the time to U = U0 /e is given by
−t
= −0.5
RC
RC
t=
.
2
(12)
(13)
(b) For a charging circuit, the
−t
q = Q0 1 − e RC ,
(14)
so the time to U = U0 /e is given by
−t
1 − e RC = e−0.5
−t
RC
(15)
−0.5
e
=1−e
−t
= ln 1 − e−0.5
RC
t = −RC ln 1 − e−0.5 = 0.933RC .
(16)
(17)
(18)
Problem 26.91. As shown in Fig. 26.83, a network of resistors of resistances R1 and R2 extends to infinity toward the
right. Prove that the total resistance RT of the infinite network is equal to
q
RT = R1 + R12 + 2R1 R2
(19)
( Hint: Since the network is infinite, the resistance of the network to the right of points c and d is also equal to RT .)
R1
R1
R1
a-/\/\/-c-/\/\/-+-/\/\/-+-...
|
|
|
Z
Z
Z
Z R2
Z R2
Z R2
Z
Z
Z
R1 |
R1 |
R1 |
b-/\/\/-d-/\/\/-+-/\/\/-+-...
Following the hint, we note that
1
1
RT = R1 +
+
R2
RT
−1
1
1
+
RT − 2R1 =
R2
RT
1
1
(RT − 2R1 ) ·
+
=1
R2
RT
RT2
−1
+ R1
(21)
(22)
(RT − 2R1 ) · (RT + R2 ) = R2 RT
(23)
− 2R1 RT + R2 RT − 2R1 R2 = R2 RT
(24)
0=
RT2
− 2R1 RT − 2R1 R2 .
Plugging this into the quadratic formula
p
q
q
2R1 ± 4R12 − 4 · 1 · (−2R1 R2 )
RT =
= R1 ± R12 + 2R1 R2 = R1 + R12 + 2R1 R2 ,
2
which is what we set out to show. Note that we chose the + case from ± because
q
R1 < R12 + 2R1 R2 ,
and RT must be greater than zero.
(20)
(25)
(26)
(27)
Problem 27.22. In an experiment with cosmic rays, a verticle beam of particles that have charge of magnitude 3e and
mass 12 times the proton mass enters a uniform horizontal magnetic field of 0.250 T and is bent in a semicircle of diameter
95.0 cm, as shown in Fig. 27.47. (a) Find the speed of the particles and the sign of their charge. (b) Is it reasonable to ignore
the gravity force on the particles? (c) How does the speed of the particles as they enter the field compare to their speed as
they exit the field?
(a) From the figure, the particle rotates counterclockwise with the B field pointing out of the page. From the righthand rule
and F = qv × B, the particle must have a negative charge. Because this centerward force makes the particle move in a circle,
v2
r
0.950 m · 3 · 1.6 · 10−19 C · 0.250 T
rqB
= 5.68 Mm/s .
=
v=
m
12 · 1.67 · 10−27 kg
Fc = qvB = mac = m
(28)
(29)
(b) The gravitational force mg = 1.97 · 10−25 N, while the magnetic force qvB = 6.82 · 10−13 N. Therefore, the gravitational
force can be safely ignored.
(c) Because the magnetic force F = qv × B is always perpendicular to the particle’s velocity v, the magnetic force will never
change the speed of a particle. Only the direction of motion is affected. So the particle leaves the field with the same speed
it entered with.
Problem 27.30. A particle with initial velocity v0 = 5.85 · 103 m/sĵ enters a region of uniform electric and magnetic fields.
The magnetic field in the region is B = −(1.35 T)k̂. Calculate the magnitude and direction of the electric field in the region
if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) −0.640 nC. You can ignore the
weight of the particle.
(a) This is a straigtforward application of the Lorentz force.
F = 0 = q(E + v × B)
(30)
E = −v × B = 5.85 · 103 m/s · 1.35 T(ĵ × k̂) = 7.90 V/mî .
(31)
(b) Because the charge canceled out in (a), the electric field should be the same: E = 7.90 V/mî.
30.0 cm
I
60.0 cm
Problem 27.35. A long wire carrying 4.50 A of current makes two 90◦ bends, as shown in Fig. 27.49. The bent part of
the wire passes through a uniform 0.240 T magnetic field direceted as shown in the figure and confined to a limited region of
space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.
60.0 cm
The verticle portion of the wire feels a force of
Fv = Ilv × B = Ilv B î = 4.50 A · 0.300 m · 0.240 Tî = 0.324 Nî .
(32)
The two horizontal legs combined feel
Fh = Ilh × B = −Ilh B ĵ = −4.50 A · 0.600 m · 0.240 Tĵ = −0.648 Nĵ .
(33)
This gives the total force a magnitude of
F =
in the direction
θ = arctan
below the horizontal.
q
Fv2 + Fh2 = 0.724 N
0.648 N
0.324 N
= arctan(2) = 63.4◦
(34)
(35)