CHAPTER 30 – Inductance; and Electromagnetic Oscillations
2.
(a) We find the mutual inductance from
M = µN1N2A/
= (2000)(4π × 10–7 T · m/A)(300 turns)(100 turns) π(0.0200 m)2/(2.44 m)
= 3.88 × 10–2 H =
38.8 mH.
(b) The induced emf in the second coil is
= – M ∆I1/∆t
4.75 V
= – (3.88 × 10–2 H)(0 – 12.0 A)/(0.0980 s) =
5.
We find the mutual inductance of the system by finding the mutual
inductance of the loop. The magnetic field of the long wire depends only
on the distance from the wire. To find the magnetic flux through the
loop, we choose a strip a distance x from the wire with width dx:
ΦΒ =
2
B · dA =
1
µ 0I
2πx
w dx =
µ0Iw
2π
ln
2
x
I
.
B
1
1
The mutual inductance is
M = ΦB/I =
(µ0w/2π) ln( 2/ 1).
7.
dx
2
We estimate the inductance by using the inductance of a solenoid:
L = µ0N2A/ = (4π × 10–7 T · m/A)(20,000 turns)2π(1.85 × 10–2 m)2/(0.45 m) =
10. We find the current from
= – L ∆I/∆t;
– 35 V = – (150 mH)(I0 – 0)/(3.0 ms), which gives I0 =
1.2 H.
0.70 A.
12. (a) The magnetic flux through coil 1 is due to its own current and the current in the other coil.
From the definition of self-inductance, the flux from its own current is Φ11 = L1I1. From the
definition of mutual inductance, the flux from the current in the other coil is Φ12 = M12I2.
Thus the total magnetic flux is
Φ1 = Φ11 + Φ12 = L1I1 + MI2.
Similarly for the other coil we have
Φ2 = Φ22 + Φ21 = L2I2 + MI1.
(b) We find the induced emf from the rate of change of the flux:
1 = – dΦ1/dt = – L1 dI1/dt – M dI2/dt.
2 = – dΦ2/dt = – L2 dI2/dt – M dI1/dt.
15. (a) For two inductors placed in series, the current through each inductor is the same. This current is
also the current through the equivalent inductor, so the total emf is
= 1+ 2
– Lseries dI/dt = (– L1 dI/dt) + (– L2 dI/dt) = – (L1 + L2) dI/dt, which gives
Lseries = L1 + L2 .
(b) For two inductors placed in parallel, the potential difference across each inductor, which is the
emf, is the same:
= 1 = 2 = – L1 dI1/dt = – L2 dI2/dt = – Lparallel dI/dt.
The total current through the equivalent inductor is
I = I1 + I2 , so we have
dI/dt = dI1/dt + dI2/dt;
– /Lparallel = – /L1 – /L2 , which gives 1/Lparallel = (1/L1) + (1/L2), or
Lparallel = L1L2/(L1 + L2).
w
19. (a) For the energy densities we have
uE = 0E2 = (8.85 × 10–12 F/m)(1.0 × 104 V/m)2 =
4.4 × 10–4 J/m3;
1.6 × 106 J/m3.
uB = B2/µ0 = (2.0 T)2/(4π × 10–7 T · m/A) =
We see that uB » uE.
(b) We find the magnitude of the electric field from
u E = 0E 2;
1.6 × 106 J/m3 = (8.85 × 10–12 F/m)E2, which gives
E = 6.0 × 108 V/m.
21. The magnetic field at the surface of the wire is
B = µ0I/2πR.
The energy density of the magnetic field is
uB = B2/µ0 = µ0I 2/8π2R2 = (4π × 10–7 T · m/A)(25 A)2/8π2(1.5 × 10–3 m)2 =
4.4 J/m3.
If V is the potential along a length of the wire, the electric field is
E = V/ = IR/ = Iρ/A = Iρ/πR2.
The energy density of the electric field is
uE = 0E2 = 0(Iρ/πR2)2
= (8.85 × 10–12 F/m)[(25 A)(1.68 × 10–8 Ω · m)/π(1.5 × 10–3 m)2]2 =
1.6 × 10–14 J/m3.
28. (a) For an LR circuit, we have
I = Imax(1 – e – t/τ);
= 1 – e – t/τ, or – t/τ = – (2.56 ms)/τ = ln , which gives τ =
3.69 ms.
(b) We find the resistance from
τ = L/R;
3.69 × 10–3 s = (310 H)/R, which gives R = 8.39 × 104 Ω =
83.9 kΩ.
30. (a) Immediately after the switch is closed, the induced emf in
L
R3
the inductor is maximum while the current in the inductor
is zero, so we have
I3
I3 = 0.
2
For loop 1, we have
R2
I1 = I2 = /(R1 + R2).
a
(b) After a long time, the currents will be constant, and there will be
no induced emf in the inductor. For the junction at point a, we have
I2
I1 = I2 + I3 .
1
For loop 1, we have
– I1R1 – I2R2 = 0.
S
For loop 2, we have
+ I2R2 – I3R3 = 0, or I2 = (R3/R2)I3.
When we use this in the junction equation, we get
I1 = [1 + (R3/R2)]I3 = [(R2 + R3)/R2]I3.
From the loop 1 equation we get
= [(R2 + R3)/R2]I3R1 + (R3/R2)I3R2 , which gives
I3 = R2 /(R1R2 + R1R3 + R2R3).
Then
I1 = (R2 + R3) /(R1R2 + R1R3 + R2R3), I2 = R3 /(R1R2 + R1R3 + R2R3).
(c) Immediately after the switch is opened, the current in R1 is zero, so we have
I1 = 0,
and the current I3 starts to decay in the LR circuit. Thus we have
– I2 = I3 = R2 /(R1R2 + R1R3 + R2R3).
(d) After a long time, all currents will be zero:
I1 = I2 = I3 = 0.
b
R1
I1
å
34. (a) The resonant frequency is
f0 = (1/2π)(1/LC)1/2 = (1/2π)[1/(175 × 10–3 H)(760 × 10–12 F)]1/2 = 1.38 × 104 Hz =
13.8 kHz.
(b) The maximum charge on the capacitor is
Q0 = CV,
so the peak value of the current is
I0 = Q0ω = CV ω = (760 × 10–12 F)(135 V)2π(1.38 × 104 Hz) = 8.90 × 10–3 A =
8.90 mA.
(c) The maximum energy stored in the inductor is
ULmax = LI02 = (175 × 10–3 H)(8.90 × 10–3 A)2 =
6.93 × 10–6 J.
CHAPTER 31 – AC Circuits
3.
6.
We find the frequency from
XC = 1/2πfC;
6.70 × 103 Ω = 1/2πf(2.40 × 10–6 F), which gives f =
We find the impedance from
Z = XL = ωL = 2πfL = 2π(33.3 kHz)(36.0 × 10–3 H) =
For the rms current we have
Irms = Vrms/XL = (750 V)/(7.53 kΩ) =
99.6 mA.
9.90 Hz.
7.53 kΩ.
9.
(a) If the voltage is
V = V0 sin ωt,
the charge on the capacitor is
ICmax
90°
Q = CV = CV0 sin ωt.
Thus the current is
I = dQ/dt = ωCV0 cos ωt = ωCV0 sin ( ωt + 90°).
(b) If the voltage is
V = V0 sin ωt,
for the circuit we have
V = V0 sin ωt = L dI/dt, or dI/dt = (V0 /L) sin ωt.
When we integrate this we get
I = – (V0/ωL) cos ωt = (V0/ ωL) sin ( ωt – 90°).
11. Because the capacitor and resistor are in parallel, their currents are
IC = V/XC , and IR = V/R.
The total current is I = IC + IR .
(a) The reactance of the capacitor is
XC1 = 1/2πf1C = 1/2π(60 Hz)(0.35 × 10–6 F) = 7.58 × 103 Ω = 7.58 kΩ.
For the fraction of current that passes through C, we have
fraction1
= IC1/(IC1 + IR) = (1/XC1)/[(1/XC1) + (1/R)] = R/(R + XC1)
= (0.400 kΩ)/(0.400 kΩ + 7.58 kΩ) = 0.050 =
5.0%.
(b) The reactance of the capacitor is
XC2 = 1/2πf2C = 1/2π(60,000 Hz)(0.35 × 10–6 F) = 7.58 Ω.
For the fraction of current that passes through C, we have
fraction2
= IC2/(IC2 + IR) = (1/XC2)/[(1/XC2) + (1/R)] = R/(R + XC2)
= (400 Ω)/(400 Ω + 7.58 Ω) = 0.98 = 98%.
Thus most of the high-frequency current passes through the capacitor.
V0
ωt
90°
ILmax
15. (a) The reactance of the capacitor is
XC = 1/2πfC = 1/2π(60 Hz)(0.80 × 10–6 F) = 3.32 × 103 Ω = 3.32 kΩ.
The impedance of the circuit is
Z = (R2 + XC2)1/2 = [(6.0 kΩ)2 + (3.32 kΩ)2]1/2 = 6.86 kΩ.
The rms current is
18 mA.
Irms = Vrms/Z = (120 V)/(6.86 kΩ) =
(b) We find the phase angle from
cos φ = R/Z = (6.0 kΩ)/(6.86 kΩ) = 0.875.
In an RC circuit, the current leads the voltage, so φ =
– 29°.
(c) The power dissipated is
P = Irms2R = (17.5 × 10–3 A)2(6.0 × 103 Ω) =
1.8 W.
(d) The rms readings across the elements are
VR = IrmsR = (17.5 mA)(6.0 kΩ) =
105 V;
VC = IrmsXC = (17.5 mA)(3.32 kΩ) =
58 V.
Note that, because the maximal voltages occur at different times, the two readings do not add
to the applied voltage of 120 V.
16. (a) The reactance of the inductor is
XL = 2πfL = 2π(60 Hz)(0.250 H) = 94.2 Ω.
The impedance of the circuit is
Z = (R2 + XL2)1/2 = [(765 Ω)2 + (94.2 Ω)2]1/2 = 770.8 Ω.
The rms current is
Irms = Vrms/Z = (120 V)/(770.8 Ω) =
0.156 A.
(b) We find the phase angle from
cos φ = R/Z = (765 Ω)/(770.8 Ω) = 0.993.
In an RL circuit, the current lags the voltage, so φ =
+ 7.02°.
(c) The power dissipated is
P = Irms2R = (0.156 A)2(765 Ω) =
18.5 W.
(d) The rms readings across the elements are
VR = IrmsR = (0.156 A)(765 Ω) =
119 V;
VL = IrmsXL = (0.156 A)(94.2 Ω) =
15 V.
Note that, because the maximal voltages occur at different times, the two readings do not add
to the applied voltage of 120 V.