1 Ω, What is the potential difference V -V

Phys102
Term: 132
Q1.
Final
Wednesday, May 21, 2014
Code: 20
Page: 1
What is the potential difference V B -V A in the circuit shown in Figure 1 if R 1 =70.0 Ω,
R 2 =105 Ω, R 3 =140 Ω, ε 1 =2.0 V and ε 2 =7.0 V?
Figure 1
A)
B)
C)
D)
E)
𝑖4
−2.3 V
+2.3 V
+3.6 V
−1.1 V
+1.1 V
𝑖2
Ans:
𝑉𝐵 − 𝑉𝐴 = −𝑖2 𝑅2
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑖2 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑜𝑜𝑝 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝜀2 − 𝑖2 (𝑅1 + 𝑅2 + 𝑅3 ) = 0
𝑖2 =
Q2.
𝜀2
7
=
= 0.022 𝐴
𝑅1 + 𝑅2 + 𝑅3
70 + 105 + 140
𝑉𝐵 − 𝑉𝐴 = −𝑖2 𝑅2 = −0.022 × 105 = −2.3 𝑉
The current in the 12- Ω resistor shown in the circuit of Figure 2 is:
Figure 2
⇒
A)
B)
C)
D)
E)
Ans:
𝑖4 = 𝑖5 =
12 V
0.50 A
1.5 A
2.5 A
2.0 A
0.30 A
𝑖0
2
12
= 2×4 =
3𝐴
2
= 1.5 𝐴
𝑇ℎ𝑒𝑛 𝑖6 × 6 = 𝑖12 × 12
𝑎𝑛𝑑 𝑖6 + 𝑖12 = 1.5 𝐴
𝑇ℎ𝑒𝑛 𝑖12 = 0.50 𝐴
12 V
4Ω
3Ω
⇓
4Ω
5Ω
4Ω
𝑖4
𝑖5
𝑖0
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 2
Q3.
A 120-V power line is protected by a 15-A fuse. What is the maximum number of
“120 V, 500 W” light bulbs, connected in parallel, that can be operated at full
brightness from this line?
A)
B)
C)
D)
E)
3
5
4
2
6
Ans:
𝑃𝑓𝑢𝑠𝑒 = 𝐼𝑉 = 15 × 120 = 1800 𝑊
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑢𝑙𝑏 𝑜𝑝𝑒𝑟𝑎𝑡𝑒𝑑 𝑁, 𝑡ℎ𝑒𝑛
Q4.
𝑁=
𝑁 ∙ 𝑃𝑏𝑢𝑙𝑏 = 𝑃𝑓𝑢𝑠𝑒
𝑃𝑓𝑢𝑠𝑒
1800
=
= 3.6 𝑏𝑢𝑙𝑏𝑠 ⇒ 𝑁 = 3
500
𝑃𝑏𝑢𝑙𝑏
Figure 3 shows three sections of a circuit that are to be connected in turn to the same
battery via a switch. The resistors as well as the capacitors are all identical. Rank the
sections according to the time required for the capacitor to reach 50% of its final
charge, greatest first.
Figure 3
i 2R
A)
B)
C)
D)
E)
1, 3, 2
2, 1, 3
3, 1, 2
1, 2, 3
2, 3, 1
Ans:
𝑡1 = −𝑅𝐶𝑙𝑛(0.5)
2
R
R/2
i
i
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 3
Q5.
ε
In Figure 4, R 1 = 5.0 Ω, R 2 = 10 Ω, R 3 = 15 Ω, C 1 = C 2 = 5.0 μF, and
= 24 V.
Assume the circuit is in the steady state. What is the power dissipated, in Watts, in the
resistor R 3 ?
Figure 4
A)
B)
C)
D)
E)
in steady
-slate
9.6
4.4
11
33
5.5
⇒
Ans:
𝑖𝑛 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑖 =
Q6.
↑o 𝜀
R2
i
R3
4Ω
i
24
𝜀
=
= 0.8 𝐴
𝑅1 + 𝑅2 + 𝑅3 5 + 10 + 15
𝑃𝑅3 = 𝑖 2 𝑅3 = (0.8)2 × 15 = 9.6 𝑊

A particle with a charge of −2.48 × 10-8 C is moving in a magnetic field B = (2.80 T)
ˆi at an instant with velocity v = (4.19 × 104 m/s) ˆi + (−3.85 × 104 m/s) ˆj ( ˆi , ˆj , and
kˆ are the unit vectors in x, y and z directions, respectively). What is the force, in
Newton, exerted on the particle by the magnetic field?
A)
B)
C)
D)
E)
Ans:
R1
−2.67 × 10-3
+2.67 × 10-3
−6.68 × 10-4
+6.68 × 10-4
−3.86 × 10-2
�⃗ �
𝐹𝐵 = 𝑞�𝑣⃗ × 𝐵
kˆ
kˆ
kˆ
kˆ
kˆ
= (−2.48 × 10−8 )(4.19 × 104 𝚤⃗ − 3.85 × 104 𝚥⃗) × (2.8 𝚤⃗)
𝐹𝐵 = 2.67 × 10−3 𝑘�
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 4
Q7.
A straight horizontal length of copper wire is located in a magnetic field of strength
0.5×10-4 T directed out of the page, as shown in the Figure 5. What is direction and
magnitude of the minimum current in the wire needed to balance the gravitational
force on the wire? [The linear density of the wire is 60.0 gram/m]
Figure 5
A)
B)
C)
D)
E)
Ans:
4
1.2×10
3.2×104
1.2×104
4.3×104
4.3×104
A, towards negative x
A, towards negative x
A, towards positive x
A, towards positive x
A, towards negative x
�⃗ � = 𝑚𝑔
𝐹𝐵 = 𝚤⃗�𝑙⃗ × 𝐵
𝑖=
FB
mg
𝑚𝑔
𝜇∙𝑙∙𝑔
𝜇𝑔
=
=
𝑙𝑏
𝑙𝐵
𝐵
60 × 10−3 × 9.8
=
= 1.176 × 104 𝐴 = 1.2 × 104 𝐴
0.5 × 10−4
Q8.
Two electrons 1 and 2 are trapped in a uniform magnetic field B and they move in a
plane perpendicular to the magnetic field in circular paths of radii R 1 and R 2 ,
respectively. The electron 1 has kinetic energy K 1 = 500 eV and electron 2 has kinetic
energy K 2 = 300 eV. What is the value of R 1 /R 2 ?
A)
B)
C)
D)
E)
Ans:
1.3
2.8
1.7
4.0
1.0
1
𝑚𝑣 2
𝑘 = 𝑚𝑣 2 𝑏𝑢𝑡
= 𝑞𝑣𝐵 ⇒ 𝑚𝑣 = 𝑅𝑞𝐵
2
𝑅
1
1 (𝑚𝑣)2 𝑅 2 𝑞 2 𝐵 2
𝑇ℎ𝑒𝑛 𝑘 = 𝑚𝑣 2 =
=
2
2 𝑚
2𝑚
𝑞12 𝑅12 𝐵12
𝑞22 𝑅22 𝐵22
𝑘1 =
𝑎𝑛𝑑 𝑘2 =
𝑏𝑢𝑡 𝐵1 = 𝐵2 , 𝑚1 = 𝑚2 , 𝑞1 = 𝑞2
2𝑚1
2𝑚2
𝑘1
𝑅12
𝑘1
500
𝑇ℎ𝑒𝑛
= 2= � = �
= 1.3
𝑘2
𝑘2
300
𝑅2
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 5
Q9.
A current I = 17.0 mA is maintained in a circular loop of 0.320 m radius which is
parallel to the y-z plane (see Figure 6). A magnetic field B = (− 0.800 kˆ ) T is
applied. Calculate the torque, in units of N.m, exerted on the loop by the magnetic
field. ( ˆi , ˆj , and kˆ are the unit vectors in x, y and z directions, respectively).
Figure 6
A) + 4.33×10-3 ˆj
B) – 4.33×10-3 ˆj
C) + 7.32×10-4 ˆj
D) – 7.32×10-4 ˆj
E) + 2.32×10-2 ˆj
Ans:
�⃗ �
�⃗ = 𝑖 �𝐴⃗ × 𝐵
�⃗ � − 𝑖𝐴𝐵�𝚤⃗ × 𝑘
𝜏⃗ = 𝜇⃗ × 𝐵
= +𝑖𝐴𝐵𝚥⃗ = +17 × 10−3 × (𝜋 × 0.322 ) × 0.8 𝚥⃗
Q10.
𝜏⃗ = 4.33 × 10−3 𝚥̂

A charged particle is projected with velocity v into a region where there exists a

uniform electric field of strength E perpendicular to a uniform magnetic field of

strength B . If the velocity of the charged particle is to remain constant, the minimum
velocity must be
A)
B)
C)
D)
E)
of magnitude E/B and perpendicular to both E and B.
of magnitude B/E and perpendicular to both E and B.
of any magnitude but at 45 degrees to both E and B.
of magnitude E/B and parallel to E.
of magnitude E/B and parallel to B.
Ans:
�⃗ , 𝐸 = 𝑣𝐵𝑠𝑖𝑛𝜃
𝑞𝐸�⃗ = 𝑞𝑣⃗ × 𝐵
|𝑣| =
|𝐸|
|𝐵|
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 6
Q11.
Figure 7 shows cross-sectional view of three long parallel straight wires held along a
circumference of a circle of radius R = 20.0 cm, located in the plane of the page. All
wires are perpendicular to the plane of the page and each wire carries 60.0 mA current
out of the page. Determine the magnitude and direction of net magnetic field at point
P at the center of the circle.
Figure 7
B1
B3
A)
B)
C)
D)
E)
Ans:
6.00 × 10-8 T
4.23 × 10-8 T
6.11 × 10-3 T
1.23 × 10-7 T
2.00 × 10-5 T
B2
along direction of
along direction of
along direction of
along direction of
along direction of
e
e
e
c
a
�⃗𝑛𝑒𝑡 = 𝐵
�⃗1 + 𝐵
�⃗2 + 𝐵
�⃗3 𝑏𝑢𝑡 𝐵
�⃗1 + 𝐵
�⃗3 = 0
𝐵
2
𝐵𝑛𝑒𝑡
Q12.
𝜇0 𝑖
4𝜋 × 10−7 × 60 × 10−3
�⃗
= 𝐵2 =
=
2𝜋𝑟
2𝜋 × 0.2
𝐵𝑛𝑒𝑡 = 6 × 10−8 𝑇
A single piece of wire carrying current i =3.2 A is bent so it to form a circular loop of
radius a, as shown in Figure 8. If magnitude of the net magnetic field at the loop
center is 5.0 × 10-5 T, determine the radius a of the circular loop.
A)
B)
C)
D)
E)
5.3 cm
7.5 cm
8.2 cm
9.1 cm
1.3 cm
Ans:
𝐵𝑛𝑒𝑡 = 𝐵𝑤𝑖𝑟𝑒 + 𝐵𝑙𝑜𝑜𝑝
𝜇0 𝑖 𝜇0 𝑖
𝜇0 𝑖 1 1
+
=
� + �
2𝜋𝑎 2𝑎
2𝑎 2𝜋 2
𝜇0 𝑖
𝐵𝑛𝑒𝑡 =
× 0.659
𝑎
𝜇0 𝑖
4𝜋 × 10−7 × 3.2
𝑎=
× 0.659 =
× 0.659
𝐵𝑛𝑒𝑡
5 × 10−5
=
𝑎 = 5.3 × 10−2 𝑚 = 5.3 𝑐𝑚
Figure 8
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 7
Q13.
Three long wires 1,2 and 3 are parallel to a z axis, and each carries a current of 2.0 A
in the positive z direction (out of page). Their point of intersection with the xy plane
form an equilateral triangle with sides of 50 cm, as shown in Figure 9. A fourth wire
(wire 4) passes through the midpoint of the base of the triangle and is parallel to the
other three wires. If the net magnetic force on wire 1 is zero, what is the magnitude
and direction of the current in wire 4?
Figure 9
𝐹14
A) 3.0 A, into the page
𝐹12
B) 3.0 A, out of the page
𝐹13
C) 6.0 A, into the page
30° 30°
D) 6.0 A, out of the page
E) 4.3 A, into the page
d
Ans:
𝑆𝑖𝑛𝑐𝑒 = 𝐹𝑛𝑒𝑡 = 0,
|𝐹14 | = (|𝐹12 | + |𝐹13 |)𝑐𝑜𝑠30
𝑎𝑛𝑑 |𝐹12 | = |𝐹13 | 𝑡ℎ𝑒𝑛
|𝐹14 | = 2|𝐹12 |𝑐𝑜𝑠𝜃
𝜇0 𝑖1 𝑖4
𝜇0 𝑖1 𝑖2
=
× 𝑐𝑜𝑠30 × 2
2𝜋 × 0.87 𝑑
2𝜋 × 𝑑
𝑖4 = 2 × 0.87 × 𝑖2 × 𝑐𝑜𝑠30 = 0.87 × 2 × 𝑐𝑜𝑠30 × 2
𝑖4 = 3.0 𝐴
0.87 d
60°
60°
d/2
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 8
Q14.
Figure 10 shows cross sectional areas of three conductors that carry current through
the plane of the Figure. The currents have the magnitude I1 =6.0 A and I 3 =2.0 A and
 
directions as shown. If the value of the line integral ∫ B.d s is +3.8 ×10-6 T.m, what
is magnitude and direction of current I 2 . The integral involves going around the path
in the counterclockwise direction, as shown in the figure.
A)
B)
C)
D)
E)
7.0 A out of the page
6.0 A into the page
5.0 A out of the page
8.0 A into the page
9.0 A out of the page
Figure 8
Ans:
� 𝑏 ∙ 𝑑𝑠 = 𝜇0 𝑖𝑛𝑒𝑡 = 𝜇0 (𝐼3 − 𝐼1 + 𝐼2 )
3.8 × 10−6 = 𝜇0 (2 − 6 + 𝐼2 ) = 𝜇0 (𝐼2 − 4)
3.8 × 10−6 = 𝜇0 (𝐼2 − 4) = 4𝜋 × 10−7 × (𝐼2 − 4)
Q15.
𝐼2 =
3.8 × 10−6
+ 4 = 3.02 + 4 = 7.0 𝐴 𝑜𝑢𝑡 𝑜𝑓 𝑝𝑎𝑔𝑒
4𝜋 × 10−7
A 1.0 m long solenoid is 10.0 cm in diameter and carries 51.9 A current to produce
0.15 Tesla magnetic field inside the solenoid (Assume solenoid to be ideal).
Determine the number of turns in the solenoid.
A)
B)
C)
D)
E)
Ans:
𝐵 = 𝜇0 ×
𝑁=
2.30 × 103
3.73 × 103
1.81 × 102
5.33 × 106
1.01 × 102
𝑁
𝐵×𝑙
×𝑖 ⇒ 𝑁 =
𝑙
𝜇0 × 𝑖 3
0.15 × 1
= 2.299 × 103 = 2.30 × 103
4𝜋 × 10−7 × 51.9
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 9
Q16.
The wire in Figure 11 carries a current I that is decreasing with time at a constant
rate. Loop A is located to the right of the wire, loop C is located to the left of the wire
and loop B is centered on the wire. The induced emf in each of the loops is such that
Figure 11
A) loop A has clockwise induced current, loop B has no
induced current, and loop C has counterclockwise
induced current
B) no emf is induced in any loop
C) All loops experience counterclockwise induced
current
D) loop A has counterclockwise induced current, loop B
has no induced current, and loop C has clockwise
induced current
E) All loops experience clockwise induced current
Ans:
Q17.
𝐴
A rectangular circuit, of area 30 cm × 60 cm, containing a 30 Ω resistor is
perpendicular to a uniform magnetic field that starts out at 3.0 T and steadily
decreases at 0.5 T/s, as shown in Figure 12. What is the current measured by the
ammeter, due to this change in magnetic field?
Figure 12
A)
B)
C)
D)
E)
Ans:
|𝜀| = 𝐴 �
3.00 mA
1.00 mA
9.00 mA
10.0 mA
1.85 mA
𝑑𝐵
�
𝑑𝑡
𝐴 = 0.3 × 0.6 = 0.18 𝑚2
𝑑𝐵
𝑏𝑢𝑡 |𝜀| = 𝐴 � 𝑑𝑡 � = 𝑖𝑅
𝑑𝐵
𝐴� �
𝑑𝑡 = 0.18 × 0.5
𝑖=
𝑅
30
𝑖 = 3 × 10−3 = 3 𝑚𝐴
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 10
Q18.
Consider a rectangular loop of width a = 20.0 cm and length b = 30.0 cm. Half the
loop is located in a region that has a magnetic field of magnitude B= 0.90 T directed
into the page, as shown in Figure 13 (Over head view). The resistance of the coil is
R= 30.0 Ω. At what rate energy is transferred to thermal energy of the loop; if it is
moved at constant speed of v = 5.00 m/s down the page.
A)
B)
C)
D)
E)
Ans:
0.027 J/s
0.003 J/s
0.810 J/s
0.337 J/s
0.006 J/s
Figure 13
𝜀 2
𝜀2
𝑃= 𝑖 𝑅= � � 𝑅=
𝑅
𝑅
2
𝑏𝑢𝑡 𝜀 = 𝐵𝑙𝑣
𝑃=
Q19.
=
𝜀2
𝐵2𝑙2𝑣 2
=
𝑅
𝑅
(0.9)2 × (0.2)2 × (5)2
= 0.027 𝐽/𝑠
𝑅
The speed of a transverse wave on a string is 200 m/s when the string tension is 150
N. What is the needed tension in order to raise the wave speed to 400 m/s?
A)
B)
C)
D)
E)
600 N
300 N
150 N
175 N
700 N
Ans:
𝜏1
𝜏2
𝑣2
𝜏2
𝑣1 = � ; 𝑣2 = � ⇒
= �
𝜇
𝜇
𝑣1
𝜏1
𝑣2 2
400 2
𝜏2 = 𝜏1 � � = 150 × �
� = 600 𝑁
𝑣1
200
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 11
Q20.
An air column 2.00 m in length is open at both ends. The frequency of a certain
harmonic is 400 Hz, and the frequency of the next higher harmonic is 500 Hz.
Determine the speed of sound in the air column.
A)
B)
C)
D)
E)
Ans:
𝑓1 =
Q21.
400 m/s
300 m/s
150 m/s
575 m/s
600 m/s
𝑣
= 𝑓𝑛+1 − 𝑓𝑛
2𝐿
𝑓1 = 500 − 400 = 100 =
𝑣
𝑣
=
⇒ 𝑣 = 100 × 4 = 400 𝑚/𝑠
2𝐿 2 × 2
What mass of steam at 100 0C must be mixed with 100 g of ice at its melting point, in
a thermally insulated container, to produce liquid water at 25 0C.[L F = 333 kJ/kg, LV =
2256 kJ/kg]
A)
B)
C)
D)
E)
17.0 g
14.8 g
9.65 g
4.64 g
29.6 g
Ans:
𝑚𝑠𝑡𝑒𝑎𝑚 �𝐿𝑣 + 𝐶𝑤 �100 − 𝑇𝑓 �� = 𝑚𝑖𝑐𝑒 �𝐶𝑤 × �𝑇𝑓 − 0� + 𝐿𝑓 �
𝑚𝑠𝑡𝑒𝑎𝑚 =
Q22.
𝑚𝑖𝑐𝑒 �𝐶𝑤 × 𝑇𝑓 + 𝐿𝑓 �
𝐿𝑣 + 𝐶𝑤 �100 − 𝑇𝑓 �
𝑚𝑠𝑡𝑒𝑎𝑚 = 0.0170 𝑘𝑔 = 17 𝑔
=
0.1(4196 × 25 + 333,000)
2,256,000 + 4196 × 75
Under constant pressure, the temperature of 3.00 mol of an ideal monatomic gas is
raised by 20.0 K. What is the change in the internal energy of the gas?
A)
B)
C)
D)
E)
Ans:
748 J
247 J
499 J
249 J
997 J
∆𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑣 ∆𝑇
3
3
= 3 × 𝑅 × ∆𝑇 = 3 × × 8.314 × 20 = 747.9 𝐽 ≈ 748 𝐽
2
2
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 12
Q23.
One mole of an ideal monatomic gas is taken through the cycle shown in Figure 14.
If T a = 590 K, T b = 450 K and T c = 300 K, calculate the efficiency of an engine
operating in this cycle.
Figure 14
A)
B)
C)
D)
E)
0.14
0.28
0.08
0.55
0.45
Ans:
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 1 −
𝑄𝐻 →
|𝑄𝐿 |
|𝑄𝐻 |
𝑏𝑢𝑡 |𝑄𝐿 | = 𝑛𝑐𝑝 ∆𝑇 = 1 ×
=
5
× 8.31 × 150 = 3116.3 𝐽
2
|𝑄𝐻 | = 𝑛𝑐𝑣 ∆𝑇 = 1 ×
Ans:
3
𝑅 × 290
2
3
× 8.31 × 290 = 3614.9 𝐽
2
3116.3
𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑐𝑦 = 1 −
= 0.14
3614.9
=
Q24.
5
𝑅 × 150
2
↑
𝑄𝐿
Two particles with charges +2 C and−4 C are arranged as shown in Figure 15. In
which region could a third particle, with charge +1 C, be placed so that the net
electrostatic force on it is zero?
Figure 15
A) I only
B) I and II only
C) III only
D) I and III only
E) II only
𝐴
Phys102
Term: 132
Q25.
Ans:
Q26.
Final
Wednesday, May 21, 2014
Code: 20
Page: 13

Figure 16 shows four possible orientations of an electric dipole p in a uniform

electric field E . Rank them according to the magnitude of the torque exerted on the
dipole by the field, least to greatest.
Figure 16
A) 1, 2 and 4
tie, then 3
45°
45°
B) 1, 2, 3, 4
C) 1, 2, 4, 3
D) 3, 2 and 4
tie, then 1
E) 4,3,2,1
𝜏⃗ = 𝑃�⃗ × 𝐸�⃗ = 𝑃𝐸 𝑠𝑖𝑛𝜃
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly
throughout its volume. At a radius r (r < R) from the center of the sphere the electric
field has a value E. If the same charge Q were distributed uniformly throughout a
sphere of radius R′ = R/2, the magnitude of the electric field at the radius r(r < R′)
would be equal to:
A)
B)
C)
D)
E)
8E
E/2
2E
E/8
4E
Ans:
𝐸(𝑟 > 𝑅) =
𝑘𝑄
𝑅3
𝑘𝑄
∙ 𝑟 𝑎𝑛𝑑 𝐸 ′ (𝑟 > 𝑅 ′ ) = (𝑅/2)3 ∙ 𝑟
𝑘𝑄
𝐸′
3 𝑟
= 𝑅
=8
𝐸 𝑟 × 𝑘𝑄
𝑅 3 /8
⇒ 𝐸 ′ = 8𝐸
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 14
Q27.
If the electric field is 15 V/m in the positive y-direction, what is the potential
difference V B −V A between points A and B? Point A is located at x = 0.0 m, y = 0.0 m
and point B is located at x = 3.0 m, y = 5.0 m.
A)
B)
C)
D)
E)
Ans:
Q28.
∆𝑉 = −𝐸�⃗ ∙ 𝑑⃗ = −𝐸𝑑𝑐𝑜𝑠𝜃 = −𝐸𝑦 𝑑𝑦
= −15 × 5 = −75 𝑉
Two electric charges Q A = + 1.0 μC and Q B = − 2.0 μC are located 0.50 m apart. How
much work is needed, by an external agent, to increase the distance between the
charges to 1.50 m?
A)
B)
C)
D)
E)
Ans:
−75 V
+75 V
+45 V
−45 V
−88 V
-3
+24 × 10 J
-3
−24 × 10 J
-3
−18 × 10 J
-3
+18 × 10 J
-3
+36 × 10 J
1 1
𝑊𝑒𝑥𝑡 = ∆𝑈 = 𝑈𝑓 − 𝑈𝑖 = 𝑘𝑄𝐴 𝑄𝐵 � − �
𝑟𝑓 𝑟𝑖
1
1
= 9 × 109 × 10−6 × (−2 × 10−6 ) �
−
�
1.5 0.5
𝑊𝑒𝑥𝑡 = 24 × 10−3 𝐽
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 15
Q29.
A parallel plate capacitor, with plate area A and 2d separation, has capacitance
.A
metallic slab has been inserted into the gap between the plates of the capacitor
halfway between the plates filling half of the gap between the plates, as shown in the
Figure 17. What is the resulting new capacitance?
Figure 17
A)
B)
C)
D)
E)
Ans:
3
Co
2
4
Co
3
3
Co
4
16
Co
9
5
Co
4
𝑁𝑒𝑤 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝐶 ′ = 𝐶1 + 𝐶2
= 𝜀0
=
𝜀0 𝐴
𝑏𝑢𝑡 𝐶0 =
𝑡ℎ𝑒𝑛
2𝑑
3 𝜀0 𝐴
3
𝐶′ = �
� = 𝐶0
2 2𝑑
2
𝐴/2
𝐴/2
+ 𝜀0
2𝑑
𝑑
𝜀0 𝐴 𝜀0 𝐴
3 𝜀0 𝐴
+
= �
�
4𝑑
2𝑑
2 2𝑑
Phys102
Term: 132
Final
Wednesday, May 21, 2014
Code: 20
Page: 16
Q30.
Figure 18 shows a current I = 1.0 A entering a truncated solid cone made of a
conducting metal. The electron drift speed at the 3.0 mm diameter end of the cone is
-4
4.0 × 10 m/s. What is the electron drift speed at the 1.5 mm diameter end of the
wire?
A)
B)
C)
D)
E)
-3
1.6 × 10
-3
1.0 × 10
-4
2.3 × 10
-5
4.4 × 10
-4
4.4 × 10
Ans:
𝐽⃗ = 𝑛𝑒𝑣𝑑 =
Figure 18
m/s
m/s
m/s
m/s
m/s
𝐼
⇒ 𝐼 = 𝑛𝑒𝑣𝑑 𝐴
𝐴
𝐼3.0 𝑚𝑚 = 𝐼1.5 𝑚𝑚 ⇒ 𝑛𝑒𝑣𝑑3.0 𝑚𝑚 × 𝐴3.0 𝑚𝑚 = 𝑛𝑒𝑣𝑑1.5 𝑚𝑚 × 𝐴1.5 𝑚𝑚
𝑣𝑑1.5 𝑚𝑚 = 𝑣𝑑3.0 𝑚𝑚 ×
𝑣𝑑1.5 𝑚𝑚 = 1.6 × 10−3 𝑚/𝑠
𝐴3.0 𝑚𝑚
=
𝐴1.5 𝑚𝑚
2
3 × 10−3
4 × 10−4 × �𝜋 × �
� �
2
2
1.5 × 10−3
𝜋×�
�
2