The long solenoid S shown (in cross section) above has 220 turns/cm and diameter
D=3.2 cm; it carries a current i=1.5A. At its center we place a 130-turn closely packed
coil C of diameter d=2.1cm. The current in the solenoid is reduced to zero at a steady
rate in t=25 ms. What is the magnitude of the emf that is induced in coil C while the
current in the solenoid is changing?
A. Coil C is in the region of magnetic field of S:
B = µ0in
B. There is a flux through each loop of coil C:
Φ1 = BA = ( µ0in)(πd 2 / 4)
C. The net flux through coil C:
Φ = NΦ1; N = 130
D. is flux is decreasing
E. Induced emf E
B f − Bi
dΦ
dB
= NA
= NA
= − NAB / t
dt
dt
t
dΦ
E=−
= NAB / t = 75mV
dt
Solenoid: n turns/cm and diameter D;
carries a current i=i0+at3. a is a constant
Coil: N turns and diameter d>D.
What is the electric field in the coil?
A. Faraday’s law:
B. Cylindrical
symmetry
 
dΦ
E = ∫ E ⋅ ds = −
dt
 
∫ E ⋅ ds =E 2πr = Eπd
r

E

E
coil
r =d/2
Solenoid: n turns/cm and diameter D;
carries a current i=i0+at3. a is a constant
Coil: N turns and diameter d>D.
What is the electric field in the coil at t=t0 sec?
solenoid
A. Faraday’s law:
B. Cylindrical
symmetry
C. Flux due to
solenoid
D. Electric field
 
dΦ
E = ∫ E ⋅ ds = −
dt
 
∫ E ⋅ ds =E 2πr = Eπd
D
coil
r =d/2
 
Φ1 = ∫ B ⋅ dA = BAsolenoid = ( µ0in)(πD 2 / 4)
coil
Φ coil = NΦ1 = N ( µ0in)(πD 2 / 4)
dΦ coil
3π 2
πD 2 di
Eπd = −
= − Nµ 0 n
= − at Nµ0 nD 2
dt
4 dt
4
Solenoid: n turns/cm and diameter D;
carries a current i=i0+at3. a is a constant
Coil: N turns and diameter d>D.
What is the electric field in the coil at t=t0 sec?
solenoid
A. Faraday’s law:
B. Cylindrical
symmetry
C. Flux due to
solenoid
D. Electric field
 
dΦ
E = ∫ E ⋅ ds = −
dt
 
∫ E ⋅ ds =E 2πr = Eπd
D
 
Φ1 = ∫ B ⋅ dA = BAsolenoid = ( µ0in)(πD 2 / 4)
coil
Φ coil = NΦ1 = N ( µ0in)(πD 2 / 4)
3
E = − at02 Nµ0 nD 2 / d
4
coil
r =d/2
What if coil is inside the solenoid? d<D
 
dΦ
E = ∫ E ⋅ ds = −
dt
 
∫ E ⋅ ds =E 2πr = Eπd
solenoid
coil
C. Flux through the coil due to solenoid
 
Φ1 = ∫ B ⋅ dA = BAcoil = ( µ0in)(πd 2 / 4) Φ = NΦ = N ( µ in)(πd 2 / 4)
coil
1
0
coil
D. Electric field
dΦ coil
3π 2
πd 2 di
Eπd = −
= − Nµ 0 n
= − at Nµ0 nd 2
dt
4 dt
4
What if coil is inside the solenoid? d<D
 
dΦ
E = ∫ E ⋅ ds = −
dt
 
∫ E ⋅ ds =E 2πr = Eπd
C. Flux through the coil due to solenoid
solenoid
coil
 
Φ1 = ∫ B ⋅ dA = BAcoil = ( µ0in)(πd 2 / 4) Φ = NΦ = N ( µ in)(πd 2 / 4)
coil
1
0
coil
D. Electric field
3
E = − at 2 Nµ0 nd
4
The figure shows three circuits with identical batteries, inductors, and resistors.
Rank the circuits according to the current through the battery (a) just after the
switch is closed and (b) a long time later, greatest first.
Right after
Long time after
A. (1)>(2)>(3)
A. (1)>(2)>(3)
B. (3)>(2)>(1)
B. (3)>(2)>(1)
C. (1)>(3)>(2)
C. (1)>(3)>(2)
D. (2)>(3)>(1)
D. (2)>(3)>(1)
E. (2)>(1)>(3)
E. (2)>(1)>(3)
F. (3)>(1)>(2)
F. (3)>(1)>(2)
The figure shows three circuits with identical batteries, inductors, and resistors.
Rank the circuits according to the current through the battery (a) just after the
switch is closed and (b) a long time later, greatest first.
Right after
No current through the coil: zero before, zero right after.
A. (1)>(2)>(3)
B. (3)>(2)>(1)
C. (1)>(3)>(2)
D. (2)>(3)>(1)
E. (2)>(1)>(3)
F. (3)>(1)>(2)
(1) i=0
(2) i=E/R
(3)i=E/2R
The figure shows three circuits with identical batteries, inductors, and resistors.
Rank the circuits according to the current through the battery (a) just after the
switch is closed and (b) a long time later, greatest first.
No resistance in the coil:
(1) i=E/2R
(2) i=2E/R
(3)i=E/R
Long time after
A. (1)>(2)>(3)
B. (3)>(2)>(1)
C. (1)>(3)>(2)
D. (2)>(3)>(1)
E. (2)>(1)>(3)
F. (3)>(1)>(2)
Example
• In the circuit shown, the switch
is in position “a” for a long
time. It is then thrown to
position “b.”
• Calculate the amplitude of the
resulting oscillating current.
1 mH
1 µF
b
10 V
a
i = −ω q0 sin(ω t + φ0 )
• Switch in position “a”: q=CV = (1 µF)(10 V) = 10 µC
• Switch in position “b”: maximum charge on C = q0 = 10 µC
• So, amplitude of oscillating current =
1
ω q0 =
(10µC) = 0.316 A
(1mH)(1µF)
Consider an LC oscillator with L=6 mH and C=1.5µF. At some moment t0
the current in the coil is i=0.1A and the voltage across the capacitor is
V=8V. Determine
a) Net energy in the oscillator;
b) Maximal current in the coil;
c) Maximal potential difference across the capacitor;
d) How long after the time t0 the current reaches the maximal amplitude
for the first time.
Consider an LC oscillator with L=6 mH and C=1.5µF. At some moment t0 the current in the
coil is i=0.1A and the voltage across the capacitor is V=8V. Determine
a) Net energy in the oscillator;
b) Maximal current in the coil;
c) Maximal potential difference across the capacitor;
d) How long after the time t0 the
2 current2reaches2 the maximal
2 amplitude for the first time.
Total energy:
Li
CV
Limax CVmax
U=
+
=
=
2
2
2
2
any time
max V
i=0
max i
V=0
a)
6 × 10−3 × 10−2 1.5 × 10−6 × 64
U=
+
= 7.8 × 10−5 J
2
2
b)
imax ≈ 0.16 A
c)
Vmax ≈ 10V
Consider an LC oscillator with L=6 mH and C=1.5µF. At some moment t0 the current in the
coil is i0=0.1A and the voltage across the capacitor is V0=8V. Determine
current
d) How long after the time t0 the current reaches the maximal amplitude for the first time.
ω=
t0
1
≈ 1.1 × 104 rad/s
LC
tmax
i = imax cos(ωt + φ ) = imax cos(ω (t − t0 ) + φ + ωt0 ) = imax cos(ω (t − t0 ) + φ0 )
At t0
i0 = imax cos φ0
To get max current
cos φ0 =
i0
imax
≈ 0.63
ω (t − t0 ) + φ0 = π
φ0 ≈ 51 ≈ 0.9rad
π − φ0
t − t0 =
≈ 2.7 × 10−4 s
ω
OR
current
d) How long after the time t0 the current reaches the maximal amplitude for the first time.
ω=
1
≈ 1.1 × 104 rad/s
LC
t0 tmax
i = imax cos(ωt + φ ) = imax cos(ω (t − t0 ) + φ + ωt0 ) = imax cos(ω (t − t0 ) + φ0 )
At t0 i0 = −imax cos φ0
To get max current
cos φ0 = −
i0
imax
≈ −0.63
ω (t − t0 ) + φ0 = π
φ0 ≈ 2.25rad
π − φ0
t − t0 =
= 8 × 10−5 s
ω
A simpler example
• In an LC circuit,
L = 40 mH; C = 4 µF
• At t = 0, the current is at maximum;
• When will the capacitor be fully
charged for the first time?
ω=
1
1
=
rad/s
−8
LC
16x10
• ω = 2500 rad/s
• T = period of one complete cycle
• T = 2π/ω = 2.5 ms
• Capacitor will be charged
after T=1/4 cycle i.e at
• t = T/4 = 0.6 ms