Homework for the week of November 24. 9th week of classes.
Ch. 30: 3, 9, 12, 21, 24, 27, 34, 37, 46, 53, 65
3.
We find the mutual inductance of the inner loop. If we assume the outer solenoid is carrying
N
current I1 , then the magnetic field inside the outer solenoid is B = µ0 1 I1. The magnetic
l
flux through each loop of the small coil is the magnetic field times the area perpendicular to
the field. The mutual inductance is given by Eq. 30-1.
NI
N 2 µ0 1 1 A2 sin θ
N1 I1
N 2 Φ 21
µ N N A sin θ
l
A2 sin θ ; M =
Φ 21 = BA2 sin θ = µ0
=
= 0 1 2 2
I1
I1
l
l
9. We draw the coil as two elements in series, and pure resistance and
R
L
a pure inductance. There is a voltage drop due to the resistance of
+
−
the coil, given by Ohm’s law, and an induced emf due to the
Einduced
I increasing
inductance of the coil, given by Eq. 30-5. Since the current is
increasing, the inductance will create a potential difference to
b
oppose the increasing current, and so there is a drop in the potential
due to the inductance. The potential difference across the coil is the sum of the two potential
drops.
dI
Vab = IR + L
= ( 3.00 A )( 3.25 Ω ) + ( 0.44 H )( 3.60 A s ) = 11.3V
dt
µ0 N 2 A µ0 N 2 π d 2
=
. The (constant) length
12. The inductance of the solenoid is given by L =
l
l
4
of the wire is given by l wire = N π d sol , and so since d sol 2 = 2.5 d sol 1 , we also know that
N1 = 2.5N 2 . The fact that the wire is tightly wound gives l sol = Nd wire . Find the ratio of the
two inductances.
N 22 2
l 2wire π 2
µ0π N 22 2
dsol 2
dsol 2
L2
Nd
N
l
l
l
4 l sol 2
=
= sol22
= 2 sol 2 2 = sol 1 = 1 wire = 1 = 2.5
2
N1 2
l wire π
L1 µ0π N1 2
l sol 2 N 2d wire N 2
dsol 1
dsol 1
4 l sol 1
l sol 1
l sol 1
21. We create an Amperean loop of radius r to calculate the magnetic field within the wire using
Eq. 28-3. Since the resulting magnetic field only depends on radius, we use Eq. 30-7 for the
energy density in the differential volume dV = 2π r l dr and integrate from zero to the radius
of the wire.
r r
µ0 Ir
 I 
2
∫ B d l = µ0 I enc → B ( 2π r ) = µ0  π R 2  (π r ) → B = 2π R 2
2
R 1  µ Ir 
µ0 I 2
U 1
0
= ∫ uB dV = ∫
2
rdr
=
π
0 2 µ  2π R 2 
l
l
4π R 4

0 
24.
(a)
constant.
0
r 3 dr =
µ0 I 2
16π
We set I equal to 75% of the maximum value in Eq. 30-9 and solve for the time
( 2.56 ms ) = 1.847 ms ≈ 1.85 ms
t
=−
ln ( 0.25 )
ln ( 0.25 )
The resistance can be calculated from the time constant using Eq. 30-10.
I = 0.75I 0 = I 0 (1 − e − t / τ ) → τ = −
(b)
∫
R
a
31.0 mH
= 16.8 Ω
τ 1.847 ms
27. (a) We use Eq. 30-5 to determine the emf in the inductor as a function of time. Since the
exponential term decreases in time, the maximum emf occurs when t = 0.
LI R
dI
d
e = − L = − L  I 0 e − tR / L  = 0 e − t / τ = V0 e− t /τ → emax = V0 .
dt
dt
L
(b)
The current is the same just before and just after the switch moves from A to B.
We use Ohm’s
law for a steady state current to determine I0 before the switch is thrown. After the
switch is thrown, the same current flows through the inductor, and therefore that current
will flow through the resistor R’. Using Kirchhoff’s loop rule we calculate the emf in
the inductor. This will be a maximum at t = 0.
R=
L
=
V0
V
 R′ 
 55R 
,
e − IR ′ = 0 → e = R′ 0 e − t / τ ′ → emax =  V0 = 
 (120 V ) = 6.6 kV
R
R
R
 R 
(a)
We calculate the resonant frequency using Eq. 30-14.
1
1
1
1
=
= 18, 450 Hz ≈ 18.5 kHz
f =
2π LC 2π ( 0.175 H ) ( 425 × 10−12 F )
I0 =
34.
(b) As shown in Eq. 30-15, we set the peak current equal to the maximum charge (from
Eq. 24-1) multiplied by the angular frequency.
I = Q0ω = CV ( 2π f ) = ( 425 × 10−12 F ) (135 V )( 2π )(18, 450 Hz )
(c)
= 6.653 × 10−3 A ≈ 6.65 mA
We use Eq. 30-6 to calculate the maximum energy stored in the inductor.
U = 12 LI 2 =
1
2
( 0.175 H ) ( 6.653 × 10−3 A )
2
= 3.87 µ J
37. As in the derivation of 30-16, we set the total energy equal to the sum of the magnetic and
electric energies, with the charge given by Eq. 30-19. We then solve for the time that the
energy is 75% of the initial energy.
Q2 − R t
Q2 − R t
Q 2 LI 2 Q02 − RL t
U = UE + UB =
+
=
e cos 2 (ω ′t + φ ) + 0 e L sin 2 (ω ′t + φ ) = 0 e L
2C
2
2C
2C
2C
2
2
Q
Q −Rt
L
L
L
0.75 0 = 0 e L → t = − ln ( 0.75 ) = − ln ( 0.75 ) ≈ 0.29
R
R
R
2C 2C
46.
(a)
Since the resistor and capacitor are in parallel, they will have the same voltage
drop across
them. We use Ohm’s law to determine the current through the resistor and Eq. 30-25 to
determine the current across the capacitor. The total current is the sum of the currents
across each element.
IR =
V
V
; IC =
= V ( 2π fC )
R
XC
(
)
( 490 Ω ) 2π ( 60 Hz ) 0.35 × 10−6 F
V ( 2π fC )
R ( 2π fC )
IC
=
=
=
I R + I C V ( 2π fC ) + V R R ( 2π fC ) + 1 ( 490 Ω ) 2π ( 60 Hz ) 0.35 × 10−6 F + 1
(
= 0.0607 ≈ 6.1%
)
(b)
We repeat part (a) with a frequency of 60,000 Hz.
(
)
( 490 Ω ) 2π ( 60,000 Hz ) 0.35 × 10−6 F
IC
=
= 0.9847 ≈ 98%
I R + I C ( 490 Ω ) 2π ( 60,000 Hz ) 0.35 × 10−6 F + 1
(
)
53. We use the rms voltage across the resistor to determine the rms current through the circuit.
Then, using the rms current and the rms voltage across the capacitor in Eq. 30-25 we
determine the frequency.
I rms =
f =
VR , rms
R
VC , rms =
I rms
2π fC
VR , rms
( 3.0 V )
I rms
=
=
= 240 Hz
−6
2π CVC , rms 2π CRVC , rms 2π 1.0 × 10 C ( 750 Ω )( 2.7 V )
(
)
Since the voltages in the resistor and capacitor are not in phase, the rms voltage across the
power source will not be the sum of their rms voltages.
65.
(a) The peak voltage across the capacitor is the peak current multiplied by the capacitive
reactance. We calculate the current in the circuit by dividing the source voltage by the
impedance, where at resonance the impedance is equal to the resistance.
V0
V
1 V0
1
VC 0 = X C I 0 =
=
= 0 T0
2π f 0 C R 2π ( RC ) f 0
2πτ
(b) We set the amplification equal to 125 and solve for the resistance.
β=
T0
1
1
1
=
→R=
=
= 130 Ω
2πτ 2π f 0 RC
2π f 0 β C 2π ( 5000 Hz )(125) 2.0 × 10−9 F
(
)