Version One – Homework 9 – Savrasov – 39821 – May 14, 2007
This print-out should have 9 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
Induced EMF for a Wire Loop
31:01, calculus, numeric, > 1 min, normal.
001 (part 1 of 4) 1 points
The circular loop of wire shown in the figure
is placed in a spatially uniform magnetic field
such that the plane of the circular loop is perpendicular to the direction for the magnetic
field as shown in the figure. The magnetic
~
field B(t)
varies with time, with the time dependence given by
B(t) = a + b t ,
where a = 0.13 T and b = 0.017 T/s.
The acceleration due to gravity is 9.8 m/s2 .
r = 2.2 cm
radius
1
Explanation:
Faraday’s Law of Induction
Eind = −
d ΦB
∆ΦB
=−
.
dt
∆t
Lenz’ Law is used to find direction of Eind .
Magnetic flux is defined by
Z
~ · dA
~ =B
~ ·A
~.
B
ΦB ≡
S
The definition of resistance is
R≡
ρL
.
A
Power dissipated in a resistor is
P = I V = I2 R =
V2
.
R
In this case, B(t) is uniform over the surface
defined by the wire loop and perpendicular to
this surface, so the magnetic flux simplifies to
ΦB (t) = B(t) A = B(t) π r 2 .
B(t)
What is the magnetic flux through the loop
as a function of time?
1. ΦB (t) = 2 a π r
2. ΦB (t) = b π r 2
3. ΦB (t) = a π r 2
4. ΦB (t) = (a + b) π r 2
³a
´
5. ΦB (t) =
+ b π r2
t
6. ΦB (t) = 2 b π r
7. ΦB (t) = (a + b t) π r 2 correct
8. ΦB (t) = 2 (a + b) π r
´
³a
+b πr
9. ΦB (t) = 2
t
10. ΦB (t) = 2 (a + b t) π r
Substituting the value of B at time, t, we find
that
ΦB (t) = (a + b t) π r 2 .
See the results of oppose the increase in the
magnetic flux, the magnetic field generated by
the induced current must be in the opposite
direction as the original magnetic field. Using
the right-hand rule, we see that this is in the
clockwise direction. Since the current must
be in the same direction as the E, the E must
be in the clockwise direction.
002 (part 2 of 4) 1 points
What is the magnitude of the induced E ?
Correct answer: 0.025849 mV.
Explanation:
Faraday’s Law tells us that the induced E
is given by
Eind = −
∆ΦB
.
∆t
The minus sign is from Lenz’s Law; it tells
us that the induced E is in the opposite direction as the direction given by applying the
Version One – Homework 9 – Savrasov – 39821 – May 14, 2007
2
right-hand rule to the direction of the original magnetic field. Plugging in our expression
for the magnetic flux, and using the condition
that the loop radius is constant, we find that
¤
£
∆ B(t) π r 2
Eind = −
∆t
dB
= −π r2
dt
2 d
(a + b t)
= −π r
dt
= −π r2 b
= −π (0.022 m)2 (0.017 T/s)
= −2.5849 × 10−5 V .
Lenz’s Law states that “The polarity of the induced E is such that
it tends to produce a current that
will create a magnetic flux to oppose the change in magnetic flux
through the loop.”
When b > 0, the magnetic field is increasing
with time. In this case, the magnetic flux is
also increasing with time. The loop area
is constant, so what will be found is that
the induced E tries to create a magnetic field
~
in the opposite direction of the B(t),
which
corresponds to an induced E in the counterclockwise direction.
The magnitude of the induced E is then
2.5849 × 10−5 V = 0.025849 mV ..
004 (part 4 of 4) 1 points
If the cross-sectional diameter (twice the radius) of the wire is 0.2 mm, and the wire is
made of a material which has a resistivity of
1.5 × 10−6 Ω m, how much power is dissipated
in the wire loop?
003 (part 3 of 4) 1 points
Assume: You are viewing the loop from the
right side, looking in the direction of the magnetic field.
What is the direction of the induced E in
the wire loop?
1. around the loop in the counter-clockwise
angular direction correct
2. around the loop in the clockwise angular
direction
Correct answer: 1.01238 × 10−10 W.
Explanation:
Electrical energy is dissipated only by resistances. Real wires always have a resistance,
though it is usually very small. We are not,
however, given the resistance, but are instead
given the resistivity. We can find the resistance from the relation,
3. radially outward
R=
4. radially inward
5. in the direction of the magnetic field
6. in the opposite direction of the magnetic
field
Explanation:
You must use Lenz’s Law to find the direction of the induced E.
r
E
ρL
,
A
were ρ is the resistivity of the material the
wire is made from (material), L is the length
of the wire, and A is the cross-sectional area of
the wire. In this case, the wire has a circular
π d2
cross-section, so A =
. Since the wire is
4
a circular loop, L = 2 π r. We know that the
power dissipated in a resistance is given by
Pdiss = I V = I 2 R =
V2
.
R
Putting this all together, we find that
B(t)
Pdiss
2
Eind
=
R
Version One – Homework 9 – Savrasov – 39821 – May 14, 2007
2
Eind
ρ L/A
E2 A
= ind
ρL
2
E π d2 /4
= ind
ρ2πr
(2.5849 × 10−5 V)2 (0.0002 m)2
=
8 (1.5 × 10−6 Ω m) (0.022 m)
=
006 (part 1 of 1) 1 points
A rectangular coil of 50 turns, 0.2 m by 0.3 m,
is rotated at 90 rad/s in a magnetic field so
that the axis of rotation is perpendicular to
the direction of the field. The maximum emf
induced in the coil is 0.5 V.
What is the magnitude of the field?
Correct answer: 1.85185 mT.
Explanation:
= 1.01238 × 10−10 W .
Let : N = 50 turns ,
ω = 90 rad/s ,
²max = 0.5 V ,
x = 0.2 m , and
y = 0.3 m .
keywords:
Lenzs Law
31:03, trigonometry, multiple choice, > 1 min,
fixed.
005 (part 1 of 1) 1 points
Lenz’s law is about the current induced in a
loop of wire when the magnetic flux through
the loop changes.
This induced current is always
3
From ²max = N A B ω, where the area A is
A = (0.2 m)(0.3 m) = 0.06 m2 ,
so
B=
²max
N Aω
0.5 V
(50 turns)(0.06 m2 )(90 rad/s)
1. clockwise.
=
2. counterclockwise.
= 0.00185185 T = 1.85185 mT .
3. in a direction opposite that of the current
producing the original flux.
4. in such a direction as to oppose the change
in the original flux. correct
5. in the same direction as that of the current
producing the original flux.
Explanation:
The direction of the induced emf and induced current can found from Lenz’s law,
which states that the polarity of the induced
emf is such that it tends to produce a current
that will create a magnetic flux to oppose the
change in magnetic flux through the loop.
keywords:
Rotated Rectangular Coil
31:04, trigonometry, numeric, > 1 min, normal.
keywords:
Series RL Circuit
32:03, trigonometry, numeric, > 1 min, normal.
007 (part 1 of 2) 1 points
The switch in a series RL circuit with a resistance of 6 Ω , inductance of 3.5 H, and voltage
of 24 V is closed at t = 0.2 s.
What is the maximum current in the circuit?
Correct answer: 4 A.
Explanation:
Let :
Imax =
E = 24 V ,
R = 6 Ω.
and
24 V
E
=
= 4A .
R
6Ω
Version One – Homework 9 – Savrasov – 39821 – May 14, 2007
4
The initial charge is
008 (part 2 of 2) 1 points
What is the current when t = 0.7 s?
Correct answer: 2.30251 A.
Explanation:
Qmax = C V
= (4 × 10−5 F) (120 V)
= 0.0048 C .
Therefore,
Let : Imax = 4 A ,
L = 3.5 H ,
R = 6 Ω.
and
The time constant of the circuit is
τ=
L
3.5 H
=
= 0.583333 s.
R
6Ω
Thus, at t = 0.7 s s the current has been
flowing for ∆t = 0.5 s and is
h
i
I = 1 − e−∆t/τ Imax
¶¸
·
µ
0.5 s
(4 A)
= 1 − exp −
0.583333 s
= 2.30251 A .
Qmax
Imax = ω Qmax = √
LC
0.0048 C
=p
(0.01 H)(4 × 10−5 F)
= 7.58947 A .
Derivation of Imax = ω Qmax :
1. Initially, there is an electric field energy
stored in the capacitor,
UC =
When the current achieves its maximum, all
the energy in the system is stored as magnetic
energy in the inductor,
UL =
keywords:
Charge and Current in LC
32:07, trigonometry, numeric, > 1 min, normal.
009 (part 1 of 1) 2 points
A 40 µF capacitor is connected in series with
a 10 mH inductance and a switch. The capacitor is first charged to a voltage of 120 V. The
charging battery is then removed. As soon as
the switch is closed, the current begins to oscillate back and forth between one direction
and the reversed direction.
What is the maximum current in the circuit?
Correct answer: 7.58947 A.
Explanation:
Let : C = 40 µF = 4 × 10−5 F ,
V = 120 V , and
L = 10 mH = 0.01 H .
1 Q2max
.
2 C
1
2
L Imax
.
2
From conservation of energy, UL = UC , we
get
Q2
2
= max .
L Imax
C
Therefore
Qmax
Imax = √
.
LC
2.
Q = Qmax cos ω t .
I=
dI
= −ω Qmax sin ω t .
dt
Therefore
Imax = ω Qmax .
keywords: