Physics 107 TUTORIAL ASSIGNMENT #9
Cutnell & Johnson, 7th edition
Chapter 20: Problems 19, 53, 61, 65, 71, 77
Chapter 21: Problems 5, 17, 35, 41, 43
Chapter 20
*19 ssm Two wires have the same cross-sectional area and are joined end to end to form a
single wire. One is tungsten, which has a temperature coefficient of resistivity of
. The other is carbon, for which
. The total resistance of
the composite wire is the sum of the resistances of the pieces. The total resistance of the
composite does not change with temperature. What is the ratio of the lengths of the tungsten and
carbon sections? Ignore any changes in length due to thermal expansion.
53 ssm Two resistors, 42.0 and
, are connected in parallel. The current through the
resistor is 3.00 A. (a) Determine the current in the other resistor. (b) What is the total
power supplied to the two resistors?
61 ssm Determine the equivalent resistance between the points A
and B for the group of resistors in the drawing.
*65 ssm The current in the
A. Find the current in (a) the
resistor.
resistor in the drawing is 0.500
resistor and in (b) the
*71 A battery delivering a current of 55.0 A to a circuit has a terminal voltage of 23.4 V. The
electric power being dissipated by the internal resistance of the battery is 34.0 W. Find the emf of
the battery.
*77 ssm Determine the voltage across the
resistor in the
drawing. Which end of the resistor is at the higher potential?
Chapter 21
5 ssm A particle with a charge of
and a speed
of 45 m/s enters a uniform magnetic field whose
magnitude is 0.30 T. For each of the cases in the
drawing, find the magnitude and direction of the
magnetic force on the particle.
17 Two isotopes of carbon, carbon-12 and carbon-13, have masses of 19.93 × 10−27 kg and
21.59 × 10−27 kg, respectively. These two isotopes are singly ionized (+e) and each is given a
speed of 6.667 × 105 m/s. The ions then enter the bending region of a mass spectrometer where
the magnetic field is 0.8500 T. Determine the spatial separation between the two isotopes after
they have traveled through a half-circle.
**35 The two conducting rails in the drawing are tilted
upward so they each make an angle of 30.0° with respect to
the ground. The vertical magnetic field has a magnitude of
0.050 T. The 0.20-kg aluminum rod (length = 1.6 m) slides
without friction down the rails at a constant velocity. How
much current flows through the bar?
41 ssm The rectangular loop in the drawing consists of 75 turns and
carries a current of I = 4.4 A. A 1.8-T magnetic field is directed along the
+y axis. The loop is free to rotate about the z axis. (a) Determine the
magnitude of the net torque exerted on the loop and (b) state whether the
35° angle will increase or decrease.
*43 Consult Interactive Solution 21.43 to see how this problem can be solved. The coil in
Figure 21-22a contains 410 turns and has an area per turn of 3.1 × 10−3 m2. The magnetic field is
0.23 T, and the current in the coil is 0.26 A.. A brake shoe is pressed perpendicularly against the
shaft to keep the coil from turning. The coefficient of static friction between the shaft and the
brake shoe is 0.76. The radius of the shaft is 0.012 m. What is the magnitude of the minimum
normal force that the brake shoe exerts on the shaft?
Solutions
Chapter 20
19. REASONING We will ignore any changes in length due to thermal expansion. Although
the resistance of each section changes with temperature, the total resistance of the composite
does not change with temperature. Therefore,
R 
R
R
R


t
u
n
g
s
t
e
n
0
c
a
r
b
o
n
0
t
u
n
g
s
t
e
n
c
a
r
b
o
n
t
t
e
m
p
e
r
a
t
u
r
e
T
A
t
r
o
o
m
t
e
m
p
e
r
a
t
u
r
e A
From Equation 20.5, we know that the temperature dependence of the resistance for a wire
of resistance Ro at temperature To is given by R = Ro[1 + (T – T0)], where  is the
temperature coefficient of resistivity. Thus,


R

R

R
(
1


T
)

R
(
1


T
)






t
u
n
g
s
t
e
n
c
a
r
b
o
n
t
u
n
g
s
t
e
n
u
n
g
s
t
e
n
0
0
0t
c
a
r
b
o
n
a
r
b
o
n
0c
Since ∆T is the same for each wire, this simplifies to
R

–
R




t
u
n
g
s
t
e
n
u
n
g
s
t
e
n
0t
c
a
r
b
o
n
c
a
r
b
o
n
0
(1)
This expression can be used to find the ratio of the resistances. Once this ratio is known, we
can find the ratio of the lengths of the sections with the aid of Equation 20.3 (L = RA/).
SOLUTION From Equation (1), the ratio of the resistances of the two sections of the wire
is
–
1
R
t
u
n
g
s
t
e
n
–
0
.
0
0
0
5
[
(
C

)
]1
0
c
a
r
b
o
n

–

–

–
1
9
R
0
.
0
0
4
5
[
(
C

)
]


t
u
n
g
s
t
e
n
c
a
r
b
o
n
  
0

Thus, using Equation 20.3, we find the ratio of the tungsten and carbon lengths to be
 
 
–
5
R
R
A
/




L


u
n
g
s
t
e
n
1
3
.
5

1
0


m


t
u
n
g
s
t
e
n0 t
u
n
g
s
t
e
nt
0
c
a
r
b
o
n




7
0




–
8


L
9
R
A
/
R
5
.
6

1
0


m






c
a
r
b
o
n
t
u
n
g
s
t
e
n
0
c
a
r
b
o
n

c
a
r
b
o
n
0
where we have used resistivity values from Table 20.1 and the fact that the two sections
have the same cross-sectional areas.
53. REASONING Since the resistors are connected in parallel, the voltage across each one is
the same and can be calculated from Ohm's Law (Equation 20.2: V=IR). Once the voltage
across each resistor is known, Ohm's law can again be used to find the current in the second
resistor. The total power consumed by the parallel combination can be found calculating the
power consumed by each resistor from Equation 20.6b: P = I 2 R. Then, the total power
consumed is the sum of the power consumed by each resistor.
SOLUTION Using data for the second resistor, the voltage across the resistors is equal to
V

I
R

(
3
.
0
0
A
)
(
6
4
.
0

)
=
1
9
2
V
a. The current through the 42.0- resistor is
I = V/R = 192V/42Ω = 4.57 A
b. The power consumed by the 42.0- resistor is
P = I2R = (4.57A)2(42) = 878 W
while the power consumed by the 64.0- resistor is
P = I2R = (3.00A)2(64) = 576 W
Therefore the total power consumed by the two resistors is 878 W + 576 W = 1454 W .
61.
SSM REASONING When two or more resistors are in series, the equivalent resistance is
given by Equation 20.16: Rs  R1  R2  R3  . . . . Likewise, when resistors are in parallel,
the expression to be solved to find the equivalent resistance is given by Equation 20.17:
1
1
1
1



 ... . We will successively apply these to the individual resistors in the
Rp R1 R2 R3
figure in the text beginning with the resistors on the right side of the figure.
SOLUTION Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance
of the combination of those two resistors is 10.0 . The 9.0- and 8.0- resistors are in
parallel; their equivalent resistance is 4.24 . The equivalent resistances of the parallel
combination (9.0  and 8.0 ) and the series combination (4.0  and the 6.0 ) are in
parallel; therefore, their equivalent resistance is 2.98 . The 2.98- combination is in
series with the 3.0- resistor, so that equivalent resistance is 5.98 . Finally, the 5.98-
combination and the 20.0- resistor are in parallel, so the equivalent resistance between the
points A and B is 4 .6  .
65.
SSM WWW REASONING Since we know that the current in the 8.00- resistor is
0.500 A, we can use Ohm's law (V = IR) to find the voltage across the 8.00- resistor. The
8.00- resistor and the 16.0- resistor are in parallel; therefore, the voltages across them
are equal. Thus, we can also use Ohm's law to find the current through the 16.0- resistor.
The currents that flow through the 8.00- and the 16.0- resistors combine to give the total
current that flows through the 20.0- resistor. Similar reasoning can be used to find the
current through the 9.00- resistor.
SOLUTION
a. The voltage across the 8.00- resistor is V8  (0.500 A)(8.00 )  4.00 V . Since this is
also the voltage that is across the 16.0- resistor, we find that the current through the
16.0- resistor is I16  (4.00 V)/(16.0  )  0.250 A . Therefore, the total current that
flows through the 20.0- resistor is
I20  0.500 A + 0.250 A = 0.750 A
b. The 8.00- and the 16.0- resistors are in parallel, so their equivalent resistance can be
1
1
1
1



 ... , and is equal to 5.33 . Therefore,
obtained from Equation 20.17,
Rp R1 R2 R3
the
equivalent
resistance
of
the
upper
branch
of
the
circuit
is
Rupper  5.33  + 20.0   25.3  , since the 5.33- resistance is in series with the 20.0-
resistance. Using Ohm's law, we find that the voltage across the upper branch must be
V  (0.750 A)(25.3 ) = 19.0 V . Since the lower branch is in parallel with the upper
branch, the voltage across both branches must be the same. Therefore, the current through
the 9.00- resistor is, from Ohm's law,
I9 
Vlower
R9

19.0 V
 2.11 A
9.00 
71. REASONING According to the discussion in Section 20.9, the emf of a battery is equal to
its terminal voltage V plus the voltage Vr across the internal resistance; Emf = V + Vr .
According to Equation 20.6a, however, the voltage across the internal resistance is related to
the current I and the power P dissipated by the internal resistance as Vr = P/ I. Thus, the emf
of the battery can be expressed as
Emf = V 
P
I
SOLUTION Using the result found above, the emf of the battery is
Emf = V 
34.0 W
P
 23.4 V 
 24.0 V
I
55.0 A
77.
SSM REASONING We begin by labeling the currents in the three resistors. The
drawing below shows the directions chosen for these currents. The directions are arbitrary,
and if any of them is incorrect, then the analysis will show that the corresponding value for
the current is negative.
We then mark the resistors with the plus and minus signs that serve as an aid in identifying
the potential drops and rises for the loop rule, recalling that conventional current is always
directed from a higher potential (+) toward a lower potential (–). Thus, given the directions
chosen for I1, I2 , and I3 , the plus and minus signs must be those shown in the drawing. We
can now use Kirchhoff's rules to find the voltage across the 5.0- resistor.
SOLUTION Applying the loop rule to the left loop (and suppressing units for onvenience)
gives
5.0 I1 10.0I 3  2.0  10.0
(1)
Similarly, for the right loop,
10.0 I 2 10.0 I 3  2.0 15.0
(2)
If we apply the junction rule to the upper junction, we obtain
I1  I 2  I 3
(3)
Subtracting Equation (2) from Equation (1) gives
5.0I1 – 10.0I2  –5.0
(4)
We now multiply Equation (3) by 10 and add the result to Equation (2); the result is
10.0I1  20.0I2  13.0
(5)
If we then multiply Equation (4) by 2 and add the result to Equation (5), we obtain
20.0I1  3.0 , or solving for I1 , we obtain I1  0.15 A . The fact that I1 is positive means that
the current in the drawing has the correct direction. The voltage across the 5.0- resistor can
be found from Ohm's law:
V  (0.15 A)(5.0 ) = 0.75 V
Current flows from the higher potential to the lower potential, and the current through the
5.0- flows from left to right, so the left end of the resistor is at the higher potential.
Chapter 21
5.
SSM REASONING AND SOLUTION The magnitude of the force can be determined
using Equation 21.1, F = q vB sin where  is the angle between the velocity and the
magnetic field. The direction of the force is determined by using Right-Hand Rule No. 1.
a. F = q vB sin 30.0° = (8.4  10–6 C)(45 m/s)(0.30 T) sin 30.0° =
5.7  10 5 N ,
directed into the paper .
b. F = q vB sin 90.0° = (8.4  10–6 C)(45 m/s)(0.30 T) sin 90.0° =
1.1  10 4 N ,
directed into the paper .
c. F = q vB sin 150° = (8.4  10–6 C)(45 m/s)(0.30 T) sin 150° =
5.7  10 5 N ,
directed into the paper .
17. REASONING The drawing shows the velocity v of
the carbon atoms as they enter the magnetic field B.
The diameter of the circular path followed by the
carbon-12 atoms is labeled as 2r12, and that of the
carbon-13 atoms as 2r13, where r denotes the radius of
the path. The radius is given by Equation 21.2 as
r  mv /  q B  , where q is the charge on the ion
(q = +e). The difference d in the diameters is
d  2r13  2r12 (see the drawing).
B (out of paper)
v
2r12
2r13
SOLUTION The spatial separation between the two isotopes after they have traveled
though a half-circle is
 m v   m v  2v
d  2r13  2r12  2  13   2  12  
 m13  m12 
 eB   eB  eB


2 6.667 105 m/s
1.60  10
19

21.59 10 kg  19.93 10 kg   1.63  10 2 m

C   0.8500 T 
_35. REASONING The following drawing shows a side view of the conducting rails and the
aluminum rod. Three forces act on the rod: (1) its weight mg, (2) the magnetic force F, and
the normal force FN. An application of Right-Hand Rule No. 1 shows that the magnetic
force is directed to the left, as shown in the drawing. Since the rod slides down the rails at a
constant velocity, its acceleration is zero. If we choose the x-axis to be along the rails,
Newton’s second law states that the net force along the x-direction is zero: Fx = max = 0.
Using the components of F and mg that are along the x-axis, Newton’s second law becomes
 F cos 30.0  mg sin 30.0  0



 Fx
The magnetic force is given by Equation 21.3 as F = ILB sin , where = 90.0is theangle
between the magnetic field and the current. We can use these two relations to find the
current in the rod.
Current (out
of paper)
FN
Aluminum
rod
F
x
mg
30.0 o
Conducting
rail
SOLUTION Substituting the expression F = ILB sin into Newton’s second law and
solving for the current I, we obtain
I 
mg sin 30.0

LB sin 90.0 cos 30.0
b
g
b0.20 kg gc9.80 m / s hsin 30.0 
b1.6 mgb0.050 Tgsin 90.0 cos 30.0
2
14 A ___
41. SSM WWW REASONING The torque on the loop is given by Equation 21.4,
  NIABsin  . From the drawing in the text, we see that the angle  between the normal to
the plane of the loop and the magnetic field is 90  35  55 . The area of the loop is
0.70 m  0.50 m = 0.35 m2.
SOLUTION
a. The magnitude of the net torque exerted on the loop is
  NIAB sin   (75)(4.4 A)(0.35 m 2 )(1.8 T) sin 55 = 170 N  m
b. As discussed in the text, when a current-carrying loop is placed in a magnetic field, the
loop tends to rotate such that its normal becomes aligned with the magnetic field. The
normal to the loop makes an angle of 55 with respect to the magnetic field. Since this angle
decreases as the loop rotates, the 35 angle increases .
43. REASONING The coil in the drawing is oriented such that the normal to the surface of the
coil is perpendicular to the magnetic field ( = 90). The magnetic torque is a maximum, and
Equation 21.4 gives its magnitude as  = NIAB sin . In this expression N is the number of
loops in the coil, I is the current, A is the area of one loop, and B is the magnitude of the
magnetic field. The torque from the brake balances this magnetic torque. The brake torque is
brake = Fbraker, where Fbrake is the brake force, and r is the radius of the shaft and also the
lever arm. The maximum value for the brake force available from static friction is
Fbrake = sFN (Equation 4.7), where FN is the normal force pressing the brake shoe against
the shaft. The maximum brake torque, then, is brake = sFNr. By setting brake = max, we will
be able to determine the magnitude of the normal force.
SOLUTION Setting the torque produced by the brake equal to the maximum torque produced by
the coil gives
 brake  
or
s FN r  NIAB sin 


–3 2
NIAB sin   410  0.26 A  3.1  10 m  0.23 T  sin 90
FN 

 8.3 N
s r
 0.76  0.012 m 