Electromagnetism quest key

Version 001 – Electromagnetism – tubman – (IBII20142015)
This print-out should have 14 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Holt SF 22C 02
001 10.0 points
The current in an AC circuit is measured with
an ammeter, which gives a reading of 8.5 A.
Calculate the maximum AC current.
1
003 (part 2 of 2) 10.0 points
Find the maximum potential difference output of the source.
Correct answer: 249 V.
Explanation:
The maximum potential difference is
Correct answer: 12.0208 A.
Explanation:
Vmax = 249 V .
Let : Irms = 8.5 A .
The rms current√is
2
Imax
Irms =
Imax = √ ,
2
2
so the maximum AC current is
√
Imax = Irms 2
√
= (8.5 A) 2
AP B 1998 MC 45
004 10.0 points
A metal spring (wrapped loosely around a
cardboard doughnut as a toroid) has its ends
attached so that it forms a circle. It is placed
in a uniform magnetic field, as shown.
= 12.0208 A .
Holt SF 22Rev 44
002 (part 1 of 2) 10.0 points
The alternating potential difference of a generator is represented by the equation
E = (249 V) sin(359 rad/s) t ,
where E is in volts and t is in seconds.
Find the frequency of the potential difference of the source.
Correct answer: 57.1366 Hz.
Explanation:
Let : E = Vmax sin ω t
= (249 V) sin(359 rad/s) t ,
ω = 2πf
= 359 rad/s .
The frequency of the source is
ω
f=
2π
359 rad/s
=
2π
= 57.1366 Hz .
B
B
B
B
Which of the following will NOT cause a
current to be induced in the spring?
and
1. Moving the spring parallel to the magnetic field correct
2. Rotating the spring about a diameter
3. Moving the spring in and out of the magnetic field
4. Changing the magnitude of the magnetic
field
Version 001 – Electromagnetism – tubman – (IBII20142015)
5. Increasing the diameter of the circle by
stretching the spring
2. clockwise. correct
Explanation:
There will be an induced current if there
is an induced emf around the spring. By
Faraday’s law, we know that
3. counter-clockwise.
E =−
AP EM 1993 FR 2 B
005 (part 1 of 3) 10.0 points
A rectangular loop of copper wire of resistance R has width a and length b. The loop
is stationary in a uniform magnetic field. The
magnetic field B at time t = 0 seconds is directed into the page as shown below. The
uniform magnetic field varies with time t according to the relationship B = B0 cos ω t,
where ω and B0 are positive constants and B
is positive when the field is directed into the
page.
B
a
b
n turns
B
Explanation:
d ΦB
,
dt
where φB is the magnetic flux through the
loop formed by the spring.
To obtain a non-zero emf, we must change
the magnetic flux. This can be accomplished
by either changing the magnitude of the magnetic field or changing the area of the loop
perpendicular to the magnetic field.
Of the five choices, only moving the spring
parallel to the magnetic field will NOT change
the magnetic flux, thus will NOT cause a
current to be induced in the spring.
B
2
π
π
When ω t = , B = B0 cos = 0; i.e.,
2
2
the field has been decreasing, and is about to
change direction. The induced current will
be in a direction to oppose this change; i.e.,
clockwise.
006 (part 2 of 3) 10.0 points
What is the expression for the magnitude of
the induced current in the loop as a function
of time in terms of a, b, B0 , ω, R, t, and
fundamental constants.
1. I =
R ω B0
| sin ω t|
ab
2. I = a b ω B0 R | sin ω t|
a b ω B0
| sin ω t| correct
R
a b B0
4. I =
| sin ω t|
R
R B0
5. I =
| sin ω t|
ab
a b B0
| sin ω t|
6. I =
ωR
Explanation:
Calculating the flux,
3. I =
B
The direction ofπthe induced current in the
loop when ω t = , after the magnetic field
2
begins to oscillate is
1. undetermined, since the current is zero.
Φ = a b B0 cos ω t .
Calculating the emf,
dΦ
(negative sign not required)
dt
= a b ω B0 sin ω t .
E =−
Version 001 – Electromagnetism – tubman – (IBII20142015)
Using Ohm’s Law, the magnitude of the current is
I
ωt
5.
I=
π
2
0
|E|
a b ω B0
=
| sin ω t| .
R
R
007 (part 3 of 3) 10.0 points
Select a graph of the induced current I vs ω t,
taking clockwise current to be positive.
3
π
3π
2
2π
5π
2
3π
I
ωt
6.
π
2
0
I
π
3π
2
2π
5π
2
3π
ωt
1.
0
π
2
π
3π
2
2π
5π
2
3π
Explanation:
The graph is a sine wave with period 2 π .
I
I
ωt
ωt
2.
0
π
2
π
3π
2
2π
5π
2
ωt
0
π
2
π
3π
2
2π
5π
2
π
2
π
3π
2
2π
5π
2
3π
3π
I
3.
0
3π
Faraday Equation
008 10.0 points
Suppose you are looking into the end of a long
cylindrical tube in which there is a uniform
magnetic field pointing away from you.
What is the direction of the induced electric
field if the magnitude of the magnetic field is
decreased with time?
1. radially inward toward the axis of the
tube
I
2. counterclockwise
ωt
4.
0
correct
π
2
π
3π
2
2π
5π
2
3π
3. toward you
4. clockwise correct
5. radially outward from the axis of the
tube
Version 001 – Electromagnetism – tubman – (IBII20142015)
4
= 312.429 N .
6. away from you
Explanation:
If the magnetic field decreases with time,
the electric field will be generated in order to
oppose the change. The right hand rule can
be applied to find that the direction of electric
field is in the clockwise direction.
Applied Force on a Bar 02
009 (part 1 of 2) 10.0 points
In the arrangement shown in the figure, the
resistor is 7 Ω and a 9 T magnetic field is
directed out of the paper. The separation
between the rails is 3 m . An applied force
moves the bar to the right at a constant speed
of 3 m/s .
To maintain the motion of the bar, a force
must be applied on the bar to balance the
magnetic force
F = FB = 312.429 N .
010 (part 2 of 2) 10.0 points
At what rate is energy dissipated in the resistor?
Correct answer: 937.286 W.
Explanation:
The power dissipated in the resistor is
P = I 2 R = (11.5714 A)2 (7 Ω)
9T
m≪1 g
7Ω
3m
I
= 937.286 W .
3 m/s
9T
Calculate the applied force required to
move the bar to the right at a constant speed
of 3 m/s. Assume the bar and rails have negligible resistance and friction. Neglect the mass
of the bar.
AP B 1998 MC 48
011 10.0 points
A single circular loop of wire in the plane of
the page is perpendicular to a uniform mag~ directed into the page, as shown.
netic field B
B
Correct answer: 312.429 N.
Explanation:
The motional emf induced in the circuit is
E = B ℓ v = (9 T) (3 m) (3 m/s)
= 81 V .
B
If the magnitude of the magnetic field is
decreasing, then the induced current in the
wire loop is
From Ohm’s law, the current flowing through
the resistor is
E
81 V
I=
=
= 11.5714 A ,
R
7Ω
so the magnitude of the force exerted on the
bar due to the magnetic field is
2. counterclockwise around the loop.
FB = I ℓ B = (11.5714 A)(3 m)(9 T)
4. directed upward out of the paper.
1. clockwise around the loop. correct
3. zero. (No current is induced.)
Version 001 – Electromagnetism – tubman – (IBII20142015)
Explanation:
Using Lenz’s law, the induced current must
be directed to counter the change of the magnetic flux through the loop; i.e., the magnetic field generated by the induced current is
pointing downward into the page. Thus, using right hand rule, the induced current must
be clockwise along the loop.
Sliding Metal Rod 03
012 (part 1 of 2) 10.0 points
The resistance of the rectangular current loop
is R, and the metal rod is sliding to the left.
The length of the rod is d, while the width of
the rails is ℓ. a and b are the contact points
where the rod touches the rails, and d > ℓ .
a
B
d
m
v
B
ℓ
b
What is the magnitude of the induced current around the loop?
1. |Iind | =
2. |Iind | =
3. |Iind | =
4. |Iind | =
5. |Iind | =
6. |Iind | =
7. |Iind | =
B d2 v
R
B d v2
R
Bdv
R2
Bℓv
correct
R
Bℓv
R2
Bdv
R
B d v2
R2
B ℓ v2
R2
B ℓ2 v
9. |Iind | =
R
B ℓ v2
10. |Iind | =
R
Explanation:
Note: The part of the rod which extends
past the rails does not have a bearing on the
answers.
Lenz’s law states that the induced current
appears such that it opposes the change in the
magnetic flux. In this case the magnetic flux
through the rectangular loop is decreasing
(since the area of the loop is decreasing) with
the direction of the flux into the page, so that
the induced magnetic field must point into the
page in order to keep the flux through the loop
constant. This corresponds to an induced current which flows in a counter-clockwise direction. Hence, if you look at the potential drop
across the resistor, then you can see that the
potential at a is greater than the potential at
b, and the direction of the induced current is
down through the metal rod.
The rate of change of the area of the rectangular loop is
8. |Iind | =
5. directed downward into the paper.
R
5
dA
dx
=ℓ
.
dt
dt
Then from Faraday’s law, the magnitude of
the induced emf is given by
E = Bℓv.
From Ohm’s law, E = I R so the magnitude
of the induced current is
Iind =
E
Bℓv
=
.
R
R
013 (part 2 of 2) 10.0 points
The magnitude of the force exerted on the
metal rod by the magnetic field, and the direction of the current through the metal bar
are given respectively by
B 2 ℓ2 v 2
~
, current flow a → b
1. kF k =
R
Version 001 – Electromagnetism – tubman – (IBII20142015)
2
~ k = B d ℓ v , current flow a → b
2. kF
R
2
2
~ k = B d v , current flow a → b
3. kF
R
2 2
~ k = B ℓ v , current flow b → a
4. kF
R
2
2
~ k = B d v , current flow b → a
5. kF
R
2
2 2
~ k = B ℓ v , current flow b → a
6. kF
R
2
B ℓ2 v
~
7. kF k =
, current flow a → b corR
rect
2 2 2
~ k = B d v , current flow a → b
8. kF
R
2
~ k = B d ℓ v , current flow b → a
9. kF
R
2
B d2 v 2
~
10. kF k =
, current flow b → a
R
Explanation:
The magnitude of the magnetic force exerted on the metal rod is given by
F = I ℓB,
where i is the induced current found in part
2.
~ = 0 outside the
Note: ℓ was used since B
rectangle.
Substituting in the expression for the induced current yields
F =
B 2 ℓ2 v
.
R
The direction of the induced current is down
(a → b) through the metal rod as explained
in Part 1.
Flat Coil of Wire
014 10.0 points
A flat coil of wire consisting of 14 turns, each
with an area of 150 cm2 , is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from
1.4 T to 14 T in 4.7 s. If the coil has a total
6
resistance of 0.2 Ω, what is the magnitude of
the induced current?
Correct answer: 2.81489 A.
Explanation:
d ΦB
Zdt
~ · dA
~
ΦB = N
B
E=−
= NBA
N A (B2 − B1 )
|E| =
t
|E|
I=
R
N A (B2 − B1 )
=
Rt
= 2.81489 A .
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