32.74. Model: Assume the battery and the connecting wires are ideal.
Visualize: Please refer to Figure P32.74.
Solve: (a) If the switch has been closed for a long time, the capacitor is fully charged and there is no current
flowing through the right branch that contains the capacitor. So, a voltage of 60 V appears across the 60 Ω
resistor and a voltage of 40 V appears across the 40 Ω resistor. That is, maximum voltage across the capacitor is
40 V. Thus, the charge on the capacitor is
Q0 = EC = (40 V)(2.0 × 10−6 F) = 80 μC
(b) Once the switch is opened, the battery is disconnected from the capacitor. The capacitor C has two
resistances (10 Ω and 40 Ω) in series and discharges according to Q = Q0e−t/RC. For Q = 0.10 Q0,
0.10Q0 = Q0e− t /(50 Ω )(2.0 μ F) ⇒ ln ( 0.10 ) = −
t
( 50 Ω )( 2.0 μ F )
⇒ t = − ( 50 Ω )( 2.0 μ F ) ln ( 0.10 ) = 0.23 ms
33.2. Model: A magnetic field is caused by an electric current.
Visualize: Please refer to Figure EX33.2.
Solve: The current in the wire is directed to the right. B2 = 20 mT + 20 mT = 40 mT because the two overlapping
wires are carrying current in the same direction and each wire produces a magnetic field having the same direction at
point 2. B3 =20 mT – 20 mT = 0 mT, because the two overlapping wires carry currents in opposite directions and
each wire produces a field having opposite directions at point 3. The currents at 4 are also in opposite directions, but
the point is to the right of one wire and to the left of the other. From the right-hand rule, the field of both currents is
out of the page. Thus B4 = 20 mT + 20 mT =40 mT.
33.6. Model: The magnetic field is that of a moving charged particle.
Visualize: Please refer to Figure EX33.6.
Solve: The Biot-Savart law is
B=
−7
−19
7
μ0 qv sin θ (10 T m/A )(1.60 × 10 C )( 2.0 × 10 m/s ) sin135°
=
= 2.83 × 10−16 T
2
2
−2
−2
4π r 2
×
+
×
2.0
10
m
2.0
10
m
(
) (
)
G
The right-hand rule for the positive charge indicates the field points out of the page. Thus, B = 2.83 × 10−16 kˆ T.
33.12. Model: The magnetic field is the superposition of the magnetic fields of three wire segments.
Visualize: Please refer to Figure EX33.12.
Solve: The magnetic field of the horizontal wire, with current I, encircles the wire. Because the dot is on the
axis of the wire, the input current creates no magnetic field at this point. The current divides at the junction, with
I/2 traveling upward and I/2 traveling downward. The right-hand rule tells us that the upward current creates a
field at the dot that is into the page; the downward current creates a field that is out of the page. Although we
could calculate the strength of each field, the symmetry of the situation (the dot is the same distance and
direction from the base of each wire) tells us that the fields of the upward and downward current must have the
G
G
same strength. Since they are in opposite directions, their sum is 0 . Altogether, then, the field at the dot is B =
G
0T.
33.14. Model: Assume the wires are infinitely long.
Visualize: Please refer to Figure EX33.14.
Solve: The magnetic field strength at point a is
G
G
G
⎛ μI
⎞
⎛μI
⎞
Bat a = Btop + Bbottom = ⎜ 0 , out of page ⎟ + ⎜ 0 , into page ⎟
⎝ 2π d
⎠ top ⎝ 2π d
⎠ bottom
⇒ Bat a =
⎞
1
1
1
μ0 I ⎛ 1
⎛
⎞
−7
−
−
⎜⎜
⎟ = ( 2 × 10 T m/A ) (10 A ) ⎜
⎟
−2
−2
2π ⎝ 2.0 cm ( 4.0 + 2.0 ) cm ⎟⎠
⎝ 2.0 × 10 m 6.0 × 10 m ⎠
G
⇒ Bat a = ( 6.7 × 10−5 T, out of page )
At points b and c,
G
⎛ μI
⎞ ⎛ μI
⎞
Bat 2 = ⎜ 0 , into page ⎟ + ⎜ 0 , into page ⎟ = ( 2.0 × 10−4 T, into page )
⎝ 2π d
⎠ ⎝ 2π d
⎠
G
⎛μI
⎞ ⎛μI
⎞
Bat 3 = ⎜ 0 , into page ⎟ + ⎜ 0 , out of page ⎟ = ( 6.7 × 10−5 T, out of page )
⎝ 2π d
⎠ ⎝ 2π d
⎠
33.18. Model: The radius of the earth is much larger than the size of the current loop.
Solve:
(a) From Equation 33.9, the magnetic field strength at the surface of the earth at the earth’s north pole is
B=
−7
22
2
μ0 μ ( 2 × 10 T m/A )( 8.0 × 10 A m )
=
= 6.2 × 10−5 T
3
6
2π z 3
6.38
×
10
m
(
)
This value is close to the value of 5 × 10−5 T given in Table 33.1.
(b) The current required to produce a dipole moment like that on the earth is
2
μ = AI = (π Rearth
) I ⇒8.0 × 1022 A m2 = π (6.38 × 106 m)2I ⇒ I = 6.3 × 108 A
Assess:
This is an extremely large current to run through a wire around the equator.
33.22. Model: Only the two currents enclosed by the closed path contribute to the line integral.
Visualize: Please refer to Figure EX33.22.
Solve: Ampere’s law gives the line integral of the magnetic field around the closed path:
G G
úB ⋅ d s = μ0 I through = 1.38 ×10−5 T m = μ0 ( I 2 + I3 ) = ( 4π × 10−7 T m/A ) (8.0 A + I3 )
1.38 × 10−5 T m
⇒ I3 = 3.0 A, out of the page.
4π × 10−7 T m/A
Assess: The right-hand rule was used above to assign a positive sign to I2. Since I3 is also positive, it is in the
same direction as I2.
⇒ ( I 3 + 8.0 A ) =
33.26. Model: A magnetic field exerts a magnetic force on a moving charge.
Visualize: Please refer to Figure EX33.26.
Solve: (a) The force on the charge is
G
G G
Fon q = qv × B = (1.60 × 10−19 C )(1.0 × 107 m/s ) cos 45°iˆ + sin 45°kˆ × 0.50iˆ T
(
= (1.60 × 10−19 C ) ( 0.50 T )
(1.0 ×10
7
2
m/s )
) (
( iˆ × iˆ + kˆ × iˆ ) = +5.7 ×10
G
G
(b) Because the cross product iˆ × iˆ in the equation for the force is zero, Fon q = 0 N.
−13
ˆj N
)
G
G
G
33.62. Model: A magnetic field exerts a magnetic force on a moving charge given by F = qv × B. Assume
the magnetic field is uniform.
Visualize: Please refer to Figure P33.62. The magnetic field points in the +z-direction. If the charged particle is
G
G
G
moving along B, F = 0 N. If v is perpendicular to B, the motion of the charged particle is a circle. However,
G
G
when v makes an angle with B, the motion of the charged particle is like a helix or a spiral. The perpendicular
component of the velocity is responsible for the circular motion, and the parallel component is responsible for the
linear motion along the magnetic field direction.
Solve: From the figure we see that v y = v cos30° and vz = v sin 30°. For the circular motion, the magnetic force
causes a centripetal acceleration. That is,
ev y B =
mv y2
r
⇒ r=
mv y
eB
( 9.11×10 kg )( 5.0 ×10 m/s ) cos30° = 0.82 mm
(1.60 ×10 C ) ( 0.030 T )
−31
=
6
−19
The time for one revolution is
T=
2π ( 8.2 × 10−4 m )
2π r
=
= 1.19 × 10−9 s
vy
( 5.0 ×106 m/s ) cos30°
The pitch p is the vertical distance covered in time T. We have
p = vzT = ( 5.0 × 106 m/s ) sin 30° (1.19 × 10−9 s ) = 3.0 × 10−3 m = 3.0 mm
33.64. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion
at constant speed.
Visualize: Please refer to Figure P33.64.
Solve: The potential difference causes an ion of mass m to accelerate from rest to a speed v. Upon entering the
magnetic field, the ion follows a circular trajectory with cyclotron radius r = mv/eB. To be detected, an ion’s
trajectory must have radius d = 2r = 8 cm. This means the ion needs the speed
v=
eBr eBd
=
m
2m
This speed was acquired by accelerating from potential V to potential 0. We can use the conservation of energy
equation to find the voltage that will accelerate the ion:
K1 + U1 = K2 + U2 ⇒ 0 J + eΔV = 12 mv 2 + 0 J ⇒ ΔV =
mv 2
2e
Using the above expression for v, the voltage that causes an ion to be detected is
2
ΔV =
mv 2 m ⎛ eBd ⎞ eB 2 d 2
= ⎜
⎟ =
2e 2e ⎝ 2m ⎠
8m
An ion’s mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example,
the mass of N +2 is
m = mN + mN – melec = 2(14.0031 u)(1.661 × 10−27 kg/u) – 9.11 × 10−31 kg = 4.65174 × 10−26 kg
Note that we’re given the atomic masses very accurately in Exercise 28. We need to retain this accuracy to tell
the difference between N +2 and CO+. The voltage for N +2 is
(1.6 ×10 C ) ( 0.200 T ) ( 0.08 m )
8 ( 4.65174 × 10 kg )
2
−19
ΔV =
= 110.07 V
Ion
Mass (kg)
N +2
4.65174 × 10−26
110.07
+
2
−26
96.36
110.11
O
+
CO
Assess:
−26
2
Accelerating voltage (V)
5.31341 × 10
4.64986 × 10−26
The difference between N +2 and CO+ is not large but is easily detectable.
33.66. Model: The loop will not rotate about the axle if the torque due to the magnetic force on the loop
balances the torque of the weight.
Visualize: Please refer to Figure P33.66.
G
Solve: The rotational equilibrium condition Στ net = 0 N m is about the axle and means that the torque from the
weight is equal and opposite to the torque from the magnetic force. We have
( 50 ×10 kg ) g ( 0.025 m ) = μ B sin 90° = ( NIA) B
( 50 × 10 kg )( 9.8 m/s ) ( 0.025 m ) = 0.123 T
⇒B=
−3
−3
2
(10 )( 2.0 A )( 0.050 m )( 0.100 m )
Assess:
The current in the loop must be clockwise for the two torques to be equal.