GenericStudent – Homework 9 – savrasov – 39823 – Jan 02, 2008
This print-out should have 12 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
001 (part 1 of 1) 1 points
A coil of 93.8 turns in the shape of a rectangle
with width 5 cm and length 10 cm is dropped
from a position where magnetic field is 0 T
to a position where the field is 0.5 T and is
directed perpendicular to the plane of the coil.
The displacement occurs in a time of 0.501 s.
Calculate the resulting average emf induced
in the coil
Correct answer: 0.468064 V.
Explanation:
Basic Concepts: Faraday’s Law of Induction:
d ΦB
E =−
dt
According to Faraday’s Law of induction,
we may work this problem out straightforwardly:
d ΦB
dt
∆ΦB
=n·
∆t
∆B · A
=n·
∆t
= 0.468064 V
|E| = n ·
Basic Concept:
Solenoid
1
Faraday’s Law for
dΦB
dt
φB = B A cos θ
θ = ωt
E = N A B ω sin(ω t) .
E = −N
Solution: For generator, the maximum emf
is
Emax = N A B ω
= N A B (2 π f )
= (893 turns)(0.16 m2 )(0.08 T)
× 2 π (52 rev/s)
= 3734.6 V .
003 (part 2 of 2) 1 points
When the maximum induced voltage occurs,
what is the orientation of the plane of the loop
with respect to the magnetic field?
1. Not enough information is given.
2. at 45 degrees to B
3. parallel to B correct
4. perpendicular to B
keywords:
002 (part 1 of 2) 1 points
A loop of area 0.16 m2 is rotating at 52 rev/s
with the axis of rotation perpendicular to a
0.08 T magnetic field.
If there are 893 turns on the loop, what is
the maximum voltage induced in it?
Correct answer: 3734.6 V.
Explanation:
Explanation:
The emf generated at time t is
E(t) = N A B ω sin(ω t)
= N A B ω sinθ
π
|E| is maximum when |sinθ| = 1, or θ = ± ,
2
so the plane of loop is parallel to B.
keywords:
Let : A = 0.16 m2 ,
f = 52 rev/s ,
B = 0.08 T , and
N = 893 turns .
004 (part 1 of 2) 1 points
A solenoid has 112 turns of wire uniformly
wrapped around an air-filled core, which has
a diameter of 12 mm and a length of 6.1 cm.
GenericStudent – Homework 9 – savrasov – 39823 – Jan 02, 2008
The permeability of free space is
1.25664 × 10−6 N/A2 .
Calculate the self-inductance of the
solenoid.
Correct answer: 2.92259 × 10−5 H.
Explanation:
2
Explanation:
Let : µ = 800 µ0 .
µ N 2 π D2
4`
µ
=
L1
µ0
= 800 (2.92259 × 10−5 H)
L2 =
Let : N = 112 ,
D = 12 mm = 0.012 m ,
` = 6.1 cm = 0.061 m , and
µ0 = 1.25664 × 10−6 N/A2 .
The self-inductance of a solenoid is given
by
NΦ
L1 =
,
I
where Φ is the total flux inside the solenoid
and I is the current in the wire wrapped
around the solenoid.
By Ampere’s Law, the magnetic field in the
solenoid is
µ0 N I
B=
,
`
where µ = µ0 is the magnetic permeability of
the air core which is the same as free space.
The magnetic flux in the solenoid is
Φ=BA=
B π D2
4
= 0.0233808 H .
keywords:
006 (part 1 of 3) 1 points
An inductor of 491 mH with a resistance of
24 Ω is connected to a power supply with a
maximum voltage of 182 V and a frequency of
62 Hz.
Find the current in the circuit.
Correct answer: 0.944118 A.
Explanation:
Let : L = 491 mH = 0.491 H ,
RL = 24 Ω ,
V = 182 V , and
f = 62 Hz .
.
Using the above expressions for Φ and B,
we obtain for the inductance of the solenoid
NBA
L1 =
I
µ0 N 2 π D 2
=
4`
= (1.25664 × 10−6 N/A2 ) (112)2
π (0.012 m)2
×
4 (0.061 m)
= 2.92259 × 10
−5
H .
005 (part 2 of 2) 1 points
The core is replaced with a soft iron rod that
has the same dimensions, but a magnetic permeability of 800 µ0 .
What is the new inductance?
Correct answer: 0.0233808 H.
You can consider this circuit as a RLC series
circuit with XC = 0 and
XL = 2 π f L
= 2 π (62 Hz) (0.491 H)
= 191.273 Ω ,
so
Z=
q
2
RL
I=
+ XL2
=
q
(24 Ω)2 + (191.273 Ω)2
= 192.773 Ω
and
182 V
V
=
= 0.944118 A .
Z
192.773 Ω
007 (part 2 of 3) 1 points
Find the phase angle between the current and
GenericStudent – Homework 9 – savrasov – 39823 – Jan 02, 2008
applied voltage.
Correct answer: 82.8482 ◦ .
Explanation:
Explanation:
Let : C = 4 µF = 4 × 10−6 F ,
L = 146 H , and
E = 9 V.
Z
In an LC circuit, the equation describing
the charge on the capacitor as a function of
time is
RL
Q = Qmax cos(ω t + δ) ,
1
where ω = √
. Since initially at t = 0, the
LC
capacitor is fully charged, Q = Qmax . This
implies that δ = 0. The first time the charge
is zero is when cos(ωt) = 0; i.e., when
π
ω t = rad
2
π
t=
2ω
π√
=
LC
2q
π
(146 H) (4 × 10−6 F)
=
2
= 0.03796 s .
φ
cos φ =
RL
Z
φ = cos
−1
µ
24 Ω
192.773 Ω
¶
= 82.8482◦ .
008 (part 3 of 3) 1 points
Find the power loss in the inductor.
Correct answer: 21.3926 W.
Explanation:
The power which is converted into joule
heat is
P = I 2 RL = (0.944118 A)2 (24 Ω)
= 21.3926 W .
keywords:
009 (part 1 of 4) 1 points
An LC circuit is shown in the figure below.
The 4 µF capacitor is initially charged by
the 9 V battery when switch S is at position
a. Then S is thrown to position b so that the
capacitor is shorted across the 146 H inductor.
146 H
4 µF
9V
3
010 (part 2 of 4) 1 points
What is the magnitude of the current at this
time?
Correct answer: 0.00148969 A.
Explanation:
The equation describing the current as a
function of time is found by differentiating (1)
with respect to time:
dQ
I=
= −ω Qmax sin(ω t) .
dt
Note that when the charge is zero on the
capacitor, the current is at its maximum level.
(Why?) Also
Qmax = V C ,
S b
a
What is the first time at which the charge
on the capacitor is zero?
Correct answer: 0.03796 s.
Hence the current Imax when the charge is
zero is
r
C
CV
Imax = ω C V = √
V
=
L
LC
r
4 × 10−6 F
=
(9 V)
146 H
= 0.00148969 A .
GenericStudent – Homework 9 – savrasov – 39823 – Jan 02, 2008
011 (part 3 of 4) 1 points
What is the energy stored in the inductor at
this time?
Correct answer: 0.000162 J.
Explanation:
The energy stored in the inductor at this
time is
1
2
L Imax
2
1
= (146 H) (0.00148969 A)2
2
= 0.000162 J .
E=
012 (part 4 of 4) 1 points
What is the energy stored in the capacitor at
this time?
Correct answer: 0 J.
Explanation:
Because the charge on the capacitor is zero,
Q2
the energy must be zero; i.e., E =
= 0.
2C
keywords:
4