Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
This print-out should have 34 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
these problem are extra credit homework,
but the material will be in the final
6.45 cm
18 cm
0.0915 A
001 (part 1 of 2) 10 points
A rectangular loop located a distance from a
long wire carrying a current is shown in the
figure. The wire is parallel to the longest side
of the loop.
4.2 cm
Find the total magnetic flux through the
loop.
Correct answer: 1.65187 × 10−9 Wb.
Explanation:
Let :
c = 6.45 cm ,
a = 4.2 cm ,
b = 18 cm , and
I = 0.0915 A .
dr
From Ampère’s law, we know that the
strength of the magnetic field created by the
current-carrying wire at a distance r from the
wire is (see figure.)
B=
µ0 I
,
2πr
so the field varies over the loop and is directed
~ is paralperpendicular to the page. Since B
~ we can express the magnetic flux
lel to dA,
through an area element dA as
Z
Φ ≡ B dA
Z
µ0 I
=
dA .
2πr
~ is not uniform but rather depends on
Note: B
r, so it cannot be removed from the integral.
In order to integrate, we express the area
element shaded in the figure as dA = b dr.
Since r is the only variable that now appears
in the integral, we obtain for the magnetic
flux
Z a+c
dr
µ0 I
b
ΦB =
2π
r
c
¯
a+c
µ0 I b
¯
ln r ¯
=
2π
µc
¶
µ0 I b
a+c
=
ln
2π
c
µ
¶
a+c
µ0 (0.0915 A)(0.18 m)
=
ln
2π
c
µ0 (0.0915 A)(0.18 m)
(0.50148)
=
2π
= 1.65187 × 10−9 Wb .
002 (part 2 of 2) 10 points
What is the direction of the magnetic field
through the rectangular loop?
r
I
1
b
1. out of the plane of the paper
c
a
2. into the plane of the paper
3. cannot be determined with information
given correct
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
Explanation:
The direction of the current is not given,
hence the absolute direction of the magnetic
field cannot be determined, although the magnetic field is perpendicular to the plane of the
paper. For example if the current flows upward (downward) the magnetic field would be
into (out of) the plane of the paper.
003 (part 1 of 1) 10 points
A coil is wrapped with 198 turns of wire on the
perimeter of a square frame of sides 34.4 cm.
Each turn has the same area, equal to that of
the frame, and the total resistance of the coil
is 1.74 Ω. A uniform magnetic field is turned
on perpendicular to the plane of the coil.
If the field changes linearly from 0 to
0.908 Wb/m2 in a time of 1.13 s, find the magnitude of the induced emf in the coil while the
field is changing.
Correct answer: 18.8274 V.
Explanation:
Basic Concept: Faraday’s Law is
E =−
d ΦB
.
dt
Solution: The magnetic flux through the
loop at t = 0 is zero since B = 0. At t =
1.13 s , the magnetic flux through the loop
is ΦB = B A = 0.107449 Wb . Therefore the
magnitude of the induced emf is
N · ∆ΦB
∆t
(198 turns) [(0.107449 Wb) − 0]
=
(1.13 s)
= 18.8274 V
|E| = 18.8274 V .
E=
Explanation:
Basic Concept:
Faraday’s Law:
d ΦB
dt
E = −N
From Faraday’s Law, we get
¯
¯
d ΦB
|E| = ¯¯ − N
dt
dB
=NA
dt
∆B
=NA
∆t
B
=NA
t
¯
¯
¯
¯
So, the time needed equals
N AB
E
(420 turns) (0.00230722 m2 ) (0.297 T)
=
(12.3 kV)
−5
= 2.33986 × 10 s .
t=
005 (part 1 of 1) 10 points
A circular conducting loop is held fixed in
a uniform magnetic field that varies in time
according to B(t) = B0 exp(−a t) where t is
in s, a is in s−1 and B is the field strength
in T at t = 0. At t = 0, the emf induced in
the loop is 0.0659 V . At t = 3.5 s, the emf is
0.0221 V, .
Find a.
Correct answer: 0.31216 s−1 .
Explanation:
Basic concept Faraday’s Law
E =A·
004 (part 1 of 1) 10 points
A magnetic field of 0.297 T exists in the region
enclosed by a solenoid that has 420 turns and
a diameter of 5.42 cm.
Within what period of time must the field
be reduced to zero if the average magnitude
of the induced emf within the coil during this
time interval is to be 12.3 kV?
Correct answer: 2.33986 × 10−5 s.
2
dB
dt
Solution: Since the emf is
E =A
dB
,
dt
since only the magnetic field is changing and
dB
= −a B exp (−at) ,
dt
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
we have 2 equations from the 2 different times.
They are
3
The power dissipated is then
V2
(0.00107819 V)2 (106 )
=
R
0.00365 Ω
= 318.49 µW
P =
and
Dividing the second equation by the first
and then taking the natural logarithm, we
have
·
¸
(0.0221 V)
− ln
(0.0659 V)
a=
= 0.31216 s−1 .
(3.5 s)
V =A
C
43.2 cm
=
.
2π
2π
A = πr = π
µ
C
2π
Y
AL
B
B
AR
¶2
B
The equations for the (right) loop CXDC
and the (left) loop CDY C are respectively
given by
1.
2.
1
=
π
Then the induced potential is
µ ¶2
1 C
dB
V =
π 2
dt
= 0.00107819 V
X
C
Hence the area is given by
2
B
dB
.
dt
The radius is found from the circumference,
(C = `), to be:
r=
i1
006 (part 1 of 1) 10 points
A(n) 43.2 cm length of wire when used as a resistor has a resistance of 0.00365 Ω. The ends
of the wire are connected to form a circular
loop, and the plane of the loop is positioned at
right angles to a uniform magnetic field that
is increasing at the rate of 0.0726 T/s.
At what rate is thermal energy generated
in the wire?
Correct answer: 318.49 µW.
Explanation:
The changing magnetic field generates a
current in the wire. The induced potential is
007 (part 1 of 3) 10 points
Assume: The induced emf for the closed
loop octagonal CXDY C is E.
A solenoid (with magnetic field B) produces a steadily increasing uniform magnetic
flux through its circular cross section. A
octagonal circuit surrounds the solenoid as
shown in the figure. The wires connecting
in the circuit are ideal, having no resistance.
The circuit consists of two identical light bulbs
(labeled X and Y ) in series. A wire connects
points C and D. The ratio of the solenoid’s
area AL left of the wire CD and the solenoid’s
AL
= 4.
area AR right of the wire CD is
AR
i2
D
i3
at t = 0 ,
−a A B = 0.0659 V ,
at t = 1.5 s ,
−a A B exp[−a (3.5 s)] = 0.0221 V .
µ ¶2
C
2
3.
4.
5.
6.
E
+ i1 R = 0 and
5
4E
− i1 R = 0 and
5
4E
− i1 R = 0 and
5
E
− i1 R = 0 and
5
4E
+ i1 R = 0 and
5
E
− i1 R = 0 and
5
4E
5
E
5
E
5
4E
5
E
5
4E
5
− i2 R = 0 .
+ i2 R = 0 .
− i2 R = 0 .
+ i2 R = 0 .
+ i2 R = 0 .
− i2 R = 0 .
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
E
7. + i1 R = 0
5
rect
and
4E
+ i2 R = 0 . cor5
4E
E
8.
+ i1 R = 0 and
− i2 R = 0 .
5
5
Explanation:
By definition, the areas of the left and right
loops are related by
AL
= 4, we can solve for AL and AR in
AR
terms of A.
Since
4A
AL =
5
A
AR = .
5
3E
correct
5R
4E
=−
5R
3E
=+
4R
E
=+
4R
E
=−
4R
3E
=+
5R
4E
=+
5R
2. i3 = −
3. i3
4. i3
5. i3
6. i3
A = A L + AR .
7. i3
8. i3
9. i3 = 0
Then we can compute the magnitude of the
induced emf around the right and left loops.
dB
A dB
1
=
= E
dt
5 dt
5
dB
4A dB
4
EL = A L
=
= E.
dt
5 dt
5
Explanation:
From the loop Eqs. 1 and 2 in Part 1, we
can solve for the currents i1 and i2 ,
ER = A R
left loop :
(1)
(2)
008 (part 2 of 3) 10 points
Note: i3 is defined as positive if it flows in the
same direction as shown in the figure.
What is the current i3 ?
1. i3 = −
3E
4R
⇒
i3 = i2 − i1 ,
we have
Since the magnetic flux is increasing, the induced emf is in the clockwise direction and the
direction of the current is counter-clockwise,
as shown in the figure.
From Kirchoff’s laws, the loop equations for
the right and left loops respectively are
right loop :
E
R
E
.
R
Since charge is conserved at a junction
i2 = i 1 + i 3
dB
dΦ
= −A
.
dt
dt
1
E + i1 R = 0
5
4
E + i2 R = 0
5
1
5
4
i2 = −
5
i1 = −
The induced emf and the changing magnetic flux are related by
E =−
4
4
5
3
=−
5
i3 = −
1 E
E
+
R 5 R
E
.
R
009 (part 3 of 3) 10 points
The ratio of the brightness of bulb Y to that
brightnessY
of bulb X,
, is
brightnessX
brightnessY
=3
1.
brightnessX
brightnessY
2.
=9
brightnessX
brightnessY
=4
3.
brightnessX
brightnessY
4.
= 16 correct
brightnessX
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
Basic Concept: Motional emf
brightnessY
5.
=2
brightnessX
E = B ` v.
Explanation:
The brightness of a bulb is proportional to
the power dissipated by it. If the resistance of
the bulb is R, then
PY
PX
2
ECDY
C
R
=
2
ECXDC
R

dB
AL

dt
=
dB
AR
dt
µ
¶2
AL
=
AR
= (4)2
= 16 .
~ = I ~` × B.
~
F
Ohm’s Law
V
.
R
Solution: The motional emf induced in the
circuit is
2
E = B `v
= (2 T) (7 m) (9 m/s)
= 126 V .


From Ohm’s law, the current flowing through
the resistor is
E
R
126 V
=
9Ω
= 14 A .
I=
m¿1 g
2T
9Ω
Magnetic force on current
I=
010 (part 1 of 2) 10 points
Given: Assume the bar and rails have negligible resistance and friction.
In the arrangement shown in the figure,
the resistor is 9 Ω and a 2 T magnetic field
is directed into the paper. The separation
between the rails is 7 m . Neglect the mass of
the bar.
An applied force moves the bar to the right
at a constant speed of 9 m/s .
7m
I
5
9 m/s
2T
Calculate the applied force required to
move the bar to the right at a constant speed
of 9 m/s.
Correct answer: 196 N.
Explanation:
Thus, the magnitude of the force exerted on
the bar due to the magnetic field is
FB = I ` B
= (14 A)(7 m)(2 T)
= 196 N .
To maintain the motion of the bar, a force
must be applied on the bar to balance the
magnetic force
F = FB
= 196 N
011 (part 2 of 2) 10 points
At what rate is energy dissipated in the resistor?
Correct answer: 1764 W.
Explanation:
The power dissipated in the resistor is
P = I2 R
= (14 A)2 (9 Ω)
= 1764 W .
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
Note: Second of four versions.
Mg
(2)
`B
To find the induced current, we use Ohm’s law
dΦ
and substitute in the induced emf, E = −
dt
I=
012 (part 1 of 4) 10 points
A bar of negligible resistance and mass m =
38 kg in the figure below is pulled horizontally
across frictionless parallel rails, also of negligible resistance, by a massless string that passes
over an ideal pulley and is attached to a suspended mass M = 210 g. The uniform magnetic field has a magnitude B = 640 mT, and
the distance between the rails is ` = 91 cm.
The rails are connected at one end by a load
resistor R = 71 mΩ. Use g = 9.8 m/s2 .
I=
dΦ
dA
=B
= B `v
dt
dt
R
B
~ g = M ~g
F
B`v
.
(5)
R
Using (2) and (5) and noting that v is the
terminal velocity v∞
B ` v∞
Mg
=
.
`B
R
d ΦB
dt
~ ·A
~
ΦB = B
E = B`v
Solution: It follows from Lenz’s law that the
magnetic force opposes the motion of the bar.
When the wire acquires steady-state speed,
the gravitational force Fg is counter-balanced
by the magnetic force Fm (see figure below)
Solving for the magnitude of the terminal
velocity v∞
013 (part 2 of 4) 10 points
What is the acceleration when the velocity
v = 0.75 s?
Correct answer: −0.0399108 m/s2 .
Explanation:
To get the velocity as a function of time we
need the acceleration
m
a(v) ≡
T
a
M
Fm
(6)
M gR
(7)
`2 B 2
(0.21 kg)(9.8 m/s2 )(0.071 Ω)
=
(0.91 m)2 (640 mT)2
= 0.430785 m/s .
E =IR=−
T
(4)
v∞ =
~ m = I ~` × B
~
F
~g − F
~m
= (M + m) ~a = F
a
(3)
I=
What is the magnitude of the terminal velocity (i.e., the eventual steady-state speed
v∞ ) reached by the bar?
Correct answer: 0.430785 m/s.
Explanation:
Basic Concepts:
~ net
F
1 dΦ
|E|
=
R
R dt
Note: We have ignored the minus sign from
the induced emf E because we will eventually evaluate the magnitude of the terminal
velocity. The flux is Φ = BA. So
m
M
6
dv
.
dt
(8)
R
Apply Newton’s second law to the bar and
the suspended mass separately
B
Fg
Fg = M g = F m = ` I B
(1)
m a = T − Fm
M a = Fg − T ,
and
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
where T is the tension in the string. Combine
and solve for a
a=
Fg − F m
,
m+M
(9)
where Fg = M g and Fm = I ` B. Further,
E
the induced current I =
and E = B ` v, so
R
I=
and
Fm =
015 (part 4 of 4) 10 points
What is the horizontal speed of the bar at
time t = 4.15907 s, assuming that the bar was
at rest at t = 0 s?
Correct answer: 0.174675 m/s.
Explanation:
Integrating (14) we have
B`v
R
B 2 `2 v
.
R
Z
(10)
ln
Since, from (1)
Fg = M g .
µ
Mg
B 2 `2
−
v,
(m + M ) R (m + M )
or
M gR
−v
2 2
dv
a=
.
= B `
R (m + M )
dt
B 2 `2
(11)
014 (part 3 of 4) 10 points
What is the time constant τ ?
Correct answer: 7.99821 s.
Explanation:
Rewriting the differential equation, Eq.
(12), in dimensionless form and isolating t
on one side and v on the other, we get
dv
dt
=
M gR
R (m + M )
−v
B 2 `2
B 2 `2
(14)
or
dv
dt
=
(15)
v∞ − v
τ
where v∞ is defined in Part 1, (7), and the
time constant τ is
R (m + M )
B 2 `2
(0.071 Ω)(38 kg + 0.21 kg)
=
(640 mT)2 (0.91 m)2
= 7.99821 s .
τ=
0
ln
dv
=
v∞ − v
µ
¶
− ln
µ
v∞ − v
v∞
¶
Z
t
0
1
v∞
dt
τ
¶
=−
=
(17)
t
−0
τ
t
τ
µ ¶
t
−
v∞ − v
τ
=e
v∞
(12)
(13)
v
1
v∞ − v
Thus, using Eqs. (9), (10), and (11), Eq. (8)
reduces to the differential equation
a=
7
v = v∞
µ ¶
t
−

τ 
1 − e




= (0.430785 m/s) 1 − e
= 0.174675 m/s .
(18)
µ
−
¶
4.15907 s
7.99821 s 

016 (part 1 of 1) 10 points
A light bulb is connected to a battery,
turned on, and is visibly lit.
An iron core is first rapidly thrust into the
coil, then rapidly withdrawn.
iron core
(16)
light bulb
S
These two actions will temporarily
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
2. have no effect on the bulb’s brightness.
3. dim the bulb both ways.
4. brighten one way, dim the other. correct
Explanation:
While the core moves, current will be induced. Moving in one direction will increase
the DC current, while moving in the other
direction will decrease the DC current.
B
B
b
L
c
v
W
017 (part 1 of 8) 10 points
A rectangular loop with resistance R has N
turns, each of length L and width W as shown
in the figure. The loop moves into a uniform
magnetic field B (into the page) with speed v.
B a
R
d
x0
0
B
a
B
c
v
b
B
0
What is
d
B
x0
d Φtotal
(the time derivative of the
dt
flux for all turns of the loop) just after the
front edge (side ab) of the loop enters the
field?
d Φtotal
= zero
dt
d Φtotal
2.
=BW v
dt
1.
d Φtotal
=NBW
dt
d Φtotal
= N BLv
4.
dt
d Φtotal
5.
= B Lv
dt
d Φtotal
6.
= N BW Lv
dt
d Φtotal
= B W Lv
7.
dt
d Φtotal
=NBL
8.
dt
d Φtotal
9.
= N B W v correct
dt
d Φtotal
10.
=NBW L
dt
Explanation:
Basic Concepts: Magnetic flux is defined
as:
~ ·A
~
Φ=B
~ · A]
~
Φtotal = N Φ = N [B
3.
1. brighten the bulb both ways.
B
8
Given:
R=the resistance of the loop,
N =the number of turns,
L=the length of each loop,
W =the width of the loop.
Find:
d Φtotal
,
(1)
dt
(2) The current in the loop,
(3) The force on the loop as it enters the field,
(4) The force on the loop as it moves within
the field.
Solution: The magnetic flux is given by
Φtotal = N (B · A) ,
where N and B are constant, but the area is
changing.
A = W (v t) initially, so
Φtotal = N B W v t
d Φtotal
dt
= N BW v.
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
018 (part 2 of 8) 10 points
d Φtotal
dΦ
Remember: E = −
= −N
.
dt
dt
What is the magnitude of the current I in
the loop just after the front edge (side ab) of
the loop enters the magnetic field?
9
What is the direction of the current I in the
loop just after it enters the magnetic field?
1. counter-clockwise correct
2. clockwise
3. No current
1. I = ER
Explanation:
The direction is counter-clockwise.
R
E
E2
3. I =
R
E2
4. I = 3
R
E
5. I = 2
R
2. I =
020 (part 4 of 8) 10 points
Given: I is the current as found in Part 2.
What is the magnitude of the force F on
the loop just after the front edge (side ab) of
the loop enters the field?
6. I = zero
7. I = ER2
R
E2
E2
9. I = 2
R
E
correct
10. I =
R
Explanation:
8. I =
E =−
d Φtotal
= −N B W v
dt
E
NBW v
=−
R
R
N Bwv
so the current I =
is in the counterR
clockwise direction (from b up to a) in order
that flux is created in the loop which opposes
the increase of flux in the loop of the uniform
magnetic field (into the page) as the loop
moves into this magnetic field.
Note: The minus sign merely indicates that
the direction of the current will be set up
in such a way so as to resist the increasing
magnetic flux.
I=
019 (part 3 of 8) 10 points
2 2
2
~ k = N B W Lv
1. kF
R2
2 2
2
~k= N B W R
2. kF
v
2
2
~k= N B W v
3. kF
R
~ k = zero
4. kF
2 2
2
~ k = N B W v correct
5. kF
R
2
2
~k= N B R
6. kF
W2 v
2 2
~ k = N B W Lv
7. kF
R
2
2
~k = N W Rv
8. kF
B2
2 2
~k= N B W
9. kF
Rv
2
2
2
~ k = N B W Lv
10. kF
R
Explanation:
The force on the loop is given by F =
N 2 B2 W 2 v
as the forces act on
NIBW =
R
the current within the field, and the horizontal
currents have equal and opposite forces. Thus
only the right hand vertical loops have a force
acting on them. This force acts to oppose the
movement of the coils, and must point right.
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
021 (part 5 of 8) 10 points
What is the direction of the force F on the
loop just after the front edge (side ab) of the
loop enters the field?
1. indeterminate, since the force is zero
2. towards the bottom of the page
3. left
4. right correct
5. towards the top of the page
Explanation:
Since
~ = W [~I × B]
~ ,
F
F̂ = Î × B̂ .
Use right hand rule of cross product, we
see that the direction of the force is pointing
towards right.
10
~ k = 0 correct
6. kF
2 2
2
~ k = N B W Lv
7. kF
R2
2
2
~k = N W Rv
8. kF
B2
2
N B2 R
~
9. kF k =
W2 v
2 2
2
~k= N B W v
10. kF
R
Explanation:
Within the field, the magnetic flux is cond Φtotal
stant, so −
= 0. Thus, E = 0, I = 0,
dt
and no force opposes the motion.
023 (part 7 of 8) 10 points
What is the direction of the force on the loop
as it moves within the field?
1. towards the top of the page
2. right
022 (part 6 of 8) 10 points
aB
Bc
v
d
B
x0
0
What is the magnitude of the force F on
the loop as it moves within the field?
b
B
~k=
1. kF
~k=
2. kF
~k=
3. kF
~k=
4. kF
~k=
5. kF
N 2 B2 W 2 R
v
N 2 B2 W v
R
2
N B2 W
Rv
2
N B2 W L v
R
2
2
N B W2 Lv
R
3. towards the bottom of the page
4. left
5. indeterminate, since the force is zero correct
Explanation:
024 (part 8 of 8) 10 points
B
a
c
B
v
b
B
d
B
x0
0
What is the direction of the force on the
loop as it moves out of the field (edge cd is in
the field while edge ab is out of the field)?
1. left
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
11
Explanation:
Given:
R = 13.1 Ω , the resistance of the loop,
N = 62 , the number of turns,
` = 9.3 m , the length of the loop, and
w = 2 m , the width of the loop.
2. right correct
3. indeterminate, since the force is zero
4. towards the top of the page
5. towards the bottom of the page
025 (part 1 of 5) 10 points
A rectangular loop with resistance 13.1 Ω
has 62 turns. The loop’s length is 9.3 m and
width is 2 m (as shown in the figure). The
loop moves with a speed of 4.1 m/s into a
region with a uniform magnetic field of 5.2 T
(into the page). The field exists in the region
0 < x < 10.6 m .
B
B a
5.2 T
B
0
9.3 m
b
c
4.1 m/s
2m
B
13.1 Ω
d
10.6 m
B
a
B
c
v
B
b
B
d
0
10.6 m
When the edge ab of the loop just enters
the field and is between 0 and 10.6 m, what is
the magnitude of the induced current in the
loop?
Correct answer: 201.808 A.
B
B a
`
v
B
B
b
R = 13.1 Ω
c
w
Explanation:
The force on the loop is given by F =
N 2 B2 W 2 v
as the forces act on
NIBW =
R
the current within the field, and the horizontal
currents have equal and opposite forces. Thus
only the right hand vertical loops have a force
acting on them. This force acts to oppose the
movement of the coils, and must point right.
d
0
10.6 m
Basic Concepts: Magnetic flux is defined
as:
~ · A)
~
Φn = N (B
Solution: We will find: (1)
d ΦN
,
dt
(2) The current in the loop,
(3) The force on the loop as it enters the field,
(4) The force on the loop as it moves into the
field.
The magnetic flux is given by ΦN = N (B ·
A) where N and B are constant, but the area
is changing. A = w (v t) initially.
ΦN = N B w v t
d ΦN
= N Bwv
dt
d ΦN
E =−
= −N B w v
dt
E
I=
R
N Bwv
=−
R
(62) (5.2 T) (2 m) (4.1 m/s)
=−
(13.1 Ω)
= −201.808 A ,
N Bwv
is in the counterR
clockwise direction (from b up to a) in order
that flux is created in the loop which opposes
the increase of flux in the loop of the uniform
magnetic field of 5.2 T (into the page) as the
loop moves into this magnetic field.
so the current I =
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
Note: The minus sign merely indicates that
the direction of the current will be set up
in such a way so as to resist the increasing
magnetic flux.
aB
12
Bc
v
026 (part 2 of 5) 10 points
When the edge ab of the loop just enters
the field and is between 0 and 10.6 m, what
force is required to keep the loop moving with
constant speed?
Correct answer: 130126 N.
Explanation:
The force on the loop is given by F =
N 2 B 2 w2 v
as the forces act on
NIBw =
R
the current within the field, and the horizontal
currents have equal and opposite forces. Thus
only the right hand vertical loops have a force
acting on them. This force acts to oppose the
movement of the coils, and must point left.
N 2 B 2 w2 v
F =
R
2
(62) (5.2 T)2 (2 m)2 (4.1 m/s)
=
(13.1 Ω)
= 130126 N .
0
10.6 m
Just after cd passes 10.6 m, while the coil is
within the region of the field, in what direction
does the current flow between a and b?
1. There is no current correct
2. b up to a
3. a down to b
Explanation:
There is no change in flux, consequently the
current is zero.
029 (part 5 of 5) 10 points
B
a
B
2. a down to b correct
3. b up to a correct
028 (part 4 of 5) 10 points
d
B
1. a down to b
2. There is no current
Explanation:
See Part 1, the current is counter-clockwise.
B
0
10.6 m
Just after edge ab exits the field, in what
direction does the current flow between a and
b?
1. b up to a
3. There is no current
c
v
b
027 (part 3 of 5) 10 points
When the edge ab of the loop just enters the
field and is between 0 and 10.6 m, what is the
direction of the induced current flow between
a and b?
d
B
b
B
Explanation:
See Part 1, the current is now clockwise;
i.e., the change in flux is opposite to that in
Part 1.
030 (part 1 of 2) 10 points
Given: g = 9.8 m/s2 .
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
4.5 m/s
sliding rod
0.42 T
`
m
sliding rod
R
v
Assume: The rod remains in contact with
the rails as it slides down the rails. The rod
experiences no friction or air drag. The rails
at each side and on the bottom have negligible
resistance.
A straight, horizontal rod slides along parallel conducting rails at an angle with the
horizontal, as shown below. The rails are connected at the bottom by a horizontal rail so
that the rod and rails forms a closed rectangular loop. A uniform vertical field exists
throughout the region.
0Ω
B
Viewed from above
1m
58 g
6.1 Ω
B
v
θ
0Ω
Viewed from the side
Viewed from above
Basic Concepts:
E =−
0.42 T
13
The movement of the rod decreases the area
of the loop, so the flux through the loop is
changing in time, and there is an induced emf
E. If we denote the area by A, this induced
emf is
/s
m
4.5
20◦
Viewed from the side
If the velocity of the rod is 4.5 m/s, what is
the current through the resistor?
Correct answer: 291.151 mA.
Explanation:
Let : ` = 1 m ,
m = 58 g ,
R = 6.1 Ω ,
v = 4.5 m/s ,
B = 0.42 T .
d ΦB
dt
E =−
dΦ
d (B A cos θ)
=−
dt
dt
dA
.
dt
since the flux is B · A = BA cos θ, where θ is
the angle between the magnetic field B and
the normal vector to the area. The magnetic
field and the angle are both constant and were
pulled out of the differentiation. Now, if we
call the distance from the rod to the resistor
x, the emf becomes
= −B cos θ
d (` x)
dt
dx
= −B ` cos θ
dt
= −B ` v cos θ .
E = −B cos θ
and
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
Thus the current in the resistor is
|E|
B ` v cos θ
=
=
R
R
(0.42 T) (1 m) (4.5 m/s) cos(20◦ )
=
(6.1 Ω)
= 0.291151 A
= 291.151 mA .
I=
031 (part 2 of 2) 10 points
What is the terminal velocity of the rod?
Correct answer: 7.61317 m/s.
Explanation:
The terminal velocity is reached when the
forces on the rod cancel, so it feels no more
acceleration. The force from the induced current is, since the rod is perpendicular to the
magnetic field,
14
terminal velocity, the power being lost in the
resistor must equal the power being gained
due to gravity. Thus
d
Ugrav
dt
E2
dz
= mg
R
dt
(B ` v0 cos θ)2
= m g v0 sin θ .
R
Solving this for v0 gives the same result as
above.
PR =
032 (part 1 of 1) 10 points
In the figure shown, the north pole of the
magnet is first moved down toward the loop
of wire, then is withdrawn upward.
N
FB,total = I ` B .
However, this force is directed parallel to the
ground. We need the component of this force
parallel to the tracks, which is
Clockwise
Counterclockwise
As viewed from above, the induced current
in the loop is
FB = I ` B cos θ .
The component of the force of gravity parallel
to the tracks is
Fg = m g sin θ .
At the terminal velocity, these forces are in
equilibrium, FB = Fg , which yields
m g sin θ =
B ` v0 cos θ
` B cos θ
R
where the expression for I from part 1 was
used. We proceed to solve for v0
R m g sin θ
[` B cos θ]2
(6.1 Ω) (0.058 kg) (9.8 m/s2 ) sin(20◦ )
=
[(1 m) (0.42 T) cos(20◦ )]2
= 7.61317 m/s .
v0 =
Rather than worrying about force components, it might be easier to use a scalar quantity, such as power. When the rod is at its
1. for both cases clockwise with increasing
magnitude.
2. for both cases counterclockwise with decreasing magnitude.
3. for both cases counterclockwise with increasing magnitude.
4. for both cases clockwise with decreasing
magnitude.
5. first clockwise, then counterclockwise.
6. first counterclockwise, then clockwise.
correct
Explanation:
From Ohm’s law and Faraday’s law, the
V
1 dΦ
current in magnitude is I =
= −
,
R
R dt
where Φ is the magnetic flux through the loop.
We know the sign of the rate of change of the
magnetic flux is changed when the magnet is
Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz
withdrawn upward, which, according to the
equation the direction of the current is also
changed.
From Lenz’s law, we know when the magnet
is moved down toward the loop, the current
in the loop is counterclockwise as viewed from
above.
033 (part 1 of 2) 10 points
The counter-clockwise circulating current in
a solenoid is increasing at a rate of 8.39 A/s.
The cross-sectional area of the solenoid is
3.14159 cm2 , and there are 163 turns on its
18.4 cm length.
What is the magnitude of the induced E
produced by the increasing current?
Correct answer: 0.478277 mV.
Explanation:
Basic Concepts:
Faraday’s Law for
solenoid
dΦ
dt
dB
= −N A
.
dt
E = −N
Magnetic field produced by the changing current is
B=
µ0 N I
L
µ0 N d I
dB
=
.
dt
L dt
Faraday’s Law for solenoid
dΦ
dt
d (BA)
= −N
dt
2
−N A
dI
=
µ0
.
L
dt
E = −N
Magnetic field induced by current
B=
µ0 N I
.
L
Solution: Thus, the induced E is
µ0 N 2 d I
A
|E| =
L
dt
15
(1.25664 × 10−6 N/A2 ) (163)2
18.4 cm
× (3.14159 cm2 )(8.39 A/s)
µ
¶µ
¶
1 m
3 mV
× 10
V
10−2 cm
= 0.478277 mV .
=
034 (part 2 of 2) 10 points
Choose the correct statement
1. The E tries to keep the current in the
solenoid flowing in the counter-clockwise direction
2. The E does not effect the current in the
solenoid
3. Not enough information is given to determine the effect of the E
4. By the right hand rule, the E produces
magnetic fields in a direction perpendicular
to the prevailing magnetic field
5. The E attempts to move the current in the
solenoid in the clockwise direction correct
Explanation:
As the current is increasing in the counterclockwise direction, by Lenz’s law, the E will
attempt to retard the current, which establishes an E that tries to counter the flow of
the current, which in this case would be in the
clockwise direction.