douglas (jed3339) – Practice Midterm 03 – yao – (54790)
This print-out should have 15 questions.
Multiple-choice questions may continue on
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before answering.
001 10.0 points
An inductor is connected to a 9.77 Hz power
supply that produces a 64.9 V rms voltage.
What minimum inductance is needed to
keep the maximum current in the circuit below 117 mA?
1. 5.87404
2. 7.31602
3. 6.71467
4. 5.68687
5. 3.56375
6. 12.7791
7. 3.23962
8. 5.29422
9. 11.2239
10. 9.77407
Two parallel wires carry equal currents in
the opposite directions. Point A is midway
between the wires, and B is an equal distance
on the other side of the wires.
A
1.
2.
Correct answer: 12.7791 H.
4.
Let : Vmax = 64.9 V and
Irms = 117 mA = 0.117 A .
The rms current is
Imax
Irms
≤ √ .
2
5.
6.
7.
8.
Since
Imax =
Vmax
Vmax
Vmax
=
=
,
XL
ωL
2πf L
Imax
Vmax
≤ √
2πf L
2
√
2 Vmax
L≥
2 π f Irms
√
2 (64.9 V)
=
2 π (9.77 Hz) (0.117 A)
= 12.7791 H .
002
10.0 points
B
What is the ratio of the magnitude of the
magnetic field at point A to that at point B?
3.
Explanation:
1
9.
10.
BA
BB
BA
BB
BA
BB
BA
BB
BA
BB
BA
BB
BA
BB
BA
BB
BA
BB
BA
BB
10
3
4
=
3
=
=2
=
1
2
=0
1
3
2
=
3
=
= 3 correct
=
5
2
=4
Explanation:
The magnetic field due to a long wire is
B=
µ0 I
.
2πr
Let the distance between the wires be r.
The magnetic field at A due to the upgoing
wire is
Bup,A =
µ0 I
µ0 I
=
.
2 π (r/2)
πr
douglas (jed3339) – Practice Midterm 03 – yao – (54790)
The right-hand rule tells us the direction is
into the paper. Due to the fact that A is
the same distance from both wires, the other
wire gives a magnetic field at A of the same
magnitude, also directed into the paper due
to the right-hand rule. The total magnetic
field at A is
2 µ0 I
BA = 2 Bup,A =
.
πr
Now, the field at B due to the upgoing wire is
µ0 I
µ0 I
Bup,B =
=
,
2 π (3 r/2)
3πr
again into the paper, while
µ0 I
µ0 I
=
Bdown,B =
2 π (r/2)
πr
out of the paper. So
BB = Bdown,B − Bup,B
µ0 I
1
=
1−
πr
3
2 µ0 I
,
=
3πr
out of the paper. Comparing magnitudes, we
find
2 µ0 I
BA
= πr = 3 .
2 µ0 I
BB
3πr
003 (part 1 of 2) 10.0 points
A positively charged particle of mass m and
charge q is accelerated from rest in the plane
of the page through a potential difference of
E between two parallel plates as shown. The
particle is injected through a hole in the righthand plate into a region containing a uniform
magnetic field of magnitude B. The particle
curves in a semicircular path of radius R and
strikes a detector.
q
Region of
Magnetic
m
Field B
hole
E
2
In which direction does the magnetic field
point?
1. out of the page correct
2. toward the top of the page
3. to the right
4. into the page
5. to the left
6. toward the bottom of the page
Explanation:
+q
m
+
−
hole
+
+
−
−
E
+
−
B
−
Because the particle curves down, the direc~ points down. By the right-hand
tion of ~v × B
~ must point out of the page .
rule, B
004 (part 2 of 2) 10.0 points
If the particle were accelerated across a potential difference of 4 E, what would be the
new radius of the semicircular path in the
magnetic field?
1. R
2. 4 R
3. R/2
4. 2 R correct
5. R/4
douglas (jed3339) – Practice Midterm 03 – yao – (54790)
Explanation:
First, the change in kinetic energy of the
charged particle is equal to the work done by
the potential difference:
r
1
2qE
2
m v = q E =⇒ v =
2
m
L=
=
005 10.0 points
What value of inductance should be used in
series with a capacitor of 1.6 pF to form an oscillating circuit that will radiate a wavelength
of 6.5 m?
1. 28.4092
2. 5.6703
3. 4.14467
4. 8.35507
5. 5.23614
6. 17.1482
7. 4.20877
8. 4.07658
9. 3.7594
10. 7.44227
Correct answer: 7.44227 µH.
Explanation:
c
2.99792 × 108 m/s
=
λ
6.5 m
= 4.61219 × 107 Hz.
The frequency is
f0 =
2π
f02
1
4 π 2 (1.6 × 10−12
006 10.0 points
Two long parallel wires carry the same
amount of current and attract each other with
F
a force per unit length of f = .
l
If both currents are halved and the wire
separation is also halved, what is the force
per unit length?
1. f
2. 8 f
f
correct
2
f
4.
4
3.
5. 4 f
6. 2 f
7.
f
8
Explanation:
Let : c = 2.99792 × 108 m/s ,
λ = 6.5 m , and
C = 1.6 pF = 1.6 × 10−12 F .
f0 =
1
4 π2 C
F)
1
×
(4.61219 × 107 Hz)2
106 µ H
×
1H
= 7.44227 µH .
Second, the radius of circular path of a
charged particle in a perpendicular magnetic
field B is given by
mv
r=
qB
Thus increasing the acceleration voltage by a
factor of 4 will cause the particle to gain twice
the speed, which results in twice the radius in
the magnetic field.
3
1
√
LC
µ0 I 1 I 2
µ0 I 2
I2
=
∝
2πr
2πr
r
2
(1/2)
f
f′ =
f=
1/2
2
f=
007 10.0 points
A negatively charged particle is moving to the
right, directly above a wire having a current
flowing to the right, as shown in the figure. In
douglas (jed3339) – Practice Midterm 03 – yao – (54790)
which direction is the magnetic force exerted
on the particle?
1. Downward
4
A rigid rectangular loop, which measures
0.30 m by 0.40 m, carries a current of 5.5 A,
as shown in the figure. A uniform external
magnetic field of magnitude 2.9 T in the negative x direction is present. Segment CD is in
the xz-plane and forms a 35◦ angle with the
z-axis, as shown. An external torque is applied to keep the loop in static equilibrium.
Find the magnitude of the external torque.
2. Out of the page
3. Upward correct
4. Into the page
5. The magnetic force is zero since the velocity is parallel to the current.
Explanation:
The magnetic field created by the wire at
the position of the particle is out of the page,
resulting in a magnetic force that is directed
in the upward direction.
008 10.0 points
The secondary coil of an ideal transformer
provides 10.0 mA at 7500 V to a neon discharge display. The primary coil operates on
120 V. What current does the primary coil
draw?
1. 1.10 N · m correct
1. 0.625 mA
2. 1.46 N · m
2. 0.160 mA
3. 0.73 N · m
3. 0.625 A correct
4. 1.91 N · m
4. 0.160 A
5. 1.57 N · m
Explanation:
For an ideal transformer, the voltages and
currents in the primary coil with N1 turns and
secondary coil with N2 turns are given by:
V2
I1
N2
=
=
.
V1
I2
N1
For V1 = 120 V, V2 = 7500 V, and I2 =
0.010 A, we have I1 = 0.625 A.
Explanation:
The external torque must be equal in magnitude to the magnetic torque on the loop,
which is given by τ = µB sin θ.
Here
τ = (5.5 A)(0.3 m)(0.4 m) is the magnetic moment of the loop, B = 2.9 T is the magnetic
field, and θ = 35◦ is the angle between the
magnetic moment and the magnetic field.
010 (part 1 of 2) 10.0 points
009
10.0 points
douglas (jed3339) – Practice Midterm 03 – yao – (54790)
ℓ
m
R
B
I
A straight rod moves along parallel conducting rails, as shown. The rails are connected at the left side through a resistor so
that the rod and rails form a closed rectangular loop. A uniform field perpendicular to the
movement of the rod exists throughout the region. Assume the rod remains in contact with
the rails as it moves. The rod experiences no
friction or air drag. The rails and rod have
negligible resistance.
5
B
From Ohm’s Law, the emf inside the loop is
1.5 m
9.4 g
6Ω
0.73 A
2.8 T
2.8 T
At what speed should the rod be moving to
produce the downward current in the resistor?
1. 0.615
2. 0.402457
3. 1.97308
4. 0.715
5. 0.848485
6. 1.04286
7. 3.12162
8. 2.14898
9. 0.427083
10. 1.35199
E =IR
and the motional emf is
E = Bℓv
E
IR
v=
=
Bℓ
Bℓ
(0.73 A) (6 Ω)
= 1.04286 m/s .
=
(2.8 T) (1.5 m)
011 (part 2 of 2) 10.0 points
What motion does the rod exhibit?
1. Moving either left or right, since both
directions produce the same result
2. Moving to the left
3. Stationary
4. Moving to the right correct
Correct answer: 1.04286 m/s.
Explanation:
Let :
I = 0.73 A ,
ℓ = 1.5 m ,
m = 9.4 g ,
R = 6 Ω , and
B = 2.8 T .
Explanation:
The downward current through the resistor reduces the existing flux ΦB in the loop;
i.e., the current produces magnetic flux in the
opposite direction as the existing flux, into
the paper. Such a response counteracts the
reduction of enclosed flux within the rail and
rod loop that arises if the rod moves to the
right, as would follow from Lenz’ law.
Therefore the rod must be moving to the
right.
012 10.0 points
Consider an electromagnetic plane wave with
a time averaged intensity 356 W/m2 .
douglas (jed3339) – Practice Midterm 03 – yao – (54790)
The speed of light is 2.99792 × 108 m/s
and the permeability of free space is 4 π ×
10−7 T·N/A .
What is the maximum magnetic field?
1. 1.12884e-06
2. 1.59116e-06
3. 1.10254e-06
4. 9.24719e-07
5. 1.80121e-06
6. 1.76833e-06
7. 1.26208e-06
8. 1.34254e-06
9. 1.52112e-06
10. 1.72757e-06
Explanation:
1. It increases by a factor of 2.
2. It remains the same.
4. It increases by a factor of 4.
Let : I = 356 W/m2 ,
µ0 = 4 π × 10−7 T·m/A ,
c = 2.99792 × 108 m/s .
5. It decreases by a factor of 4.
and
The average intensity of an electromagnetic
wave is the magnitude of its average Poynting
vector
I = Save
Emax Bmax
EB
=
=
µ0
2µ0
Explanation:
The energy stored in an inductor is given
by
2
1 2 1
V
U = LI = L
,
2
2
R
so if both L and R are doubled, the energy
decreases by a factor of 2.
014 10.0 points
A circular coil is made of N turns of copper wire as shown in the figure. A resistor
R is inserted in the copper wire. Initially, a
uniform external magnetic field points horizontally from left to right through the coil.
Since E = c B,
2
Bmax
c
2µ
r 0
2 µ0 I
=
q c
I=
=
the energy stored in the inductor after the
switch has been closed for a long time?
3. It decreases by a factor of 2. correct
Correct answer: 1.72757 × 10−6 T.
Bmax
6
2 (4 π × 10−7 T·N/A)
s
356 W/m2
×
2.99792 × 108 m/s
= 1.72757 × 10
−6
T .
013 10.0 points
An inductor with inductance L and a resistor
with resistance R are connected in series to a
DC voltage source. If both the resistance and
the inductance are doubled, what happens to
R
Magnetic
Field B(t)
During a time interval t, the external magnetic field first decreases to zero at a a constant rate, then increases from zero at the
same constant rate, but in the opposite direction, until a reversed field (i.e., from right to
douglas (jed3339) – Practice Midterm 03 – yao – (54790)
left) equal in magnitude to the initial field is
reached.
What is the direction of the induced current
in the resistor R?
1. There is no induced current.
2. It first flows from left to right, and then
from right to left.
3. It first flows from right to left, and then
from left to right.
4. It flows from right to left.
5. It flows from left to right. correct
Explanation:
As the left-to-right magnetic field decreases
(and eventually flipping sign and increasing
in magnitude) it follows from Lenz’s law (opposition to the change in magnetic field will
tend to keep the current constant and flowing in the same direction) that the induced
emf will produce a left-to-right magnetic field
arising from induced currents in the coil.
By the right hand rule, the induced current
flows counter-clockwise when viewed from
the right and the coils are wound counterclockwise as the wire goes from the right to
left terminals. The current must enter the
loop from the right terminal and exit at the
left terminal.
Since the current is continuous, the current
must flow through the resistor in the left-toright direction.
015 10.0 points
An unpolarized light beam with intensity of
I0 passes through 2 polarizers shown in the
picture.
Unpolarized
light
Polarizer
E0
Analyzer
θ
Transmission
axis
E 0 cos θ
Polarized
lihgt
If θ = 30◦ ,what is the beam intensity after
7
the second polarizer?
5
I0
16
1
2. I2 = I0
16
9
3. I2 = I0
16
1
4. I2 = I0
2
5
5. I2 = I0
8
1
6. I2 = I0
4
3
7. I2 = I0
16
1
8. I2 = I0
8
3
9. I2 = I0 correct
8
7
10. I2 = I0
16
Explanation:
The beam intensity after the first polarizer
is
I0
E0 =
.
2
We use the formula for the intensity of the
transmitted (polarized) light. Thus the beam
intensity after the second polarizer is
1. I2 =
I = E0 cos2 θ
I0
=
cos2 (30◦ )
2
3 I0
=
8