Download Exam3_T133

Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 1
Q1.
A string, of length 0.75 m and fixed at both ends, is vibrating in its fundamental mode.
The maximum transverse speed and magnitude of maximum transverse acceleration of a
point on the string are 4.2 m/s and 840 m/s2, respectively. What is the speed of waves on
the string?
A)
B)
C)
D)
E)
48 m/s
17 m/s
22 m/s
66 m/s
53 m/s
Ans:
𝑣 = 𝜆𝑓 𝑏𝑢𝑡 𝜆 = 2 × 𝐿 = 2 × 0.75 = 1.5 𝑚
𝑓=
Q2.
𝜔
1 𝑎𝑚𝑎𝑥
84.0 1
=
�
�=
×
= 31.83 𝐻𝑧
2𝜋
2𝜋 𝑢𝑚𝑎𝑥
4.2 2𝜋
𝑣 = 𝜆𝑓 = 1.5 × 31.83 = 47.8 𝑚/𝑠
A sound wave is travelling in air. The intensity and frequency of the wave are both
doubled. What is the ratio of the new pressure amplitude to the initial pressure
amplitude?
A)
B)
C)
D)
E)
2
1/ 2
1
2
1/2
Ans:
2
𝑃𝑚𝑎𝑥
= 2𝐼𝜌𝑣 = 2𝐼𝜌𝜆𝑓 𝑏𝑢𝑡 𝑣 = 𝜆1 𝑓1 = 𝜆2 𝑓2 ; 𝑓2 = 2𝑓1 𝑡ℎ𝑒𝑛 𝜆2 =
2
𝑃1−𝑚𝑎𝑥
= 2𝐼1 𝜌𝜆1 𝑓1
2
𝑃2−𝑚𝑎𝑥
= 2𝐼2 𝜌2𝜆1
𝑓1
= 2 × 2𝐼1 𝜌𝜆1 𝑓1 = 4𝐼1 𝜌𝜆1 𝑓1
2
2
2
𝑃2−𝑚𝑎𝑥
4𝐼1 𝜌𝜆1 𝑓1
𝑃2−𝑚𝑎𝑥
=
=
2
⇒
= √2
2
2𝐼1 𝜆1 𝑓1
𝑃1−𝑚𝑎𝑥
𝑃1−𝑚𝑎𝑥
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
𝜆1
2
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 2
Q3.
A copper container with a mass of 0.500 kg contains 0.180 kg of water, and both are at a
temperature of 20.0 oC. A 0.250 kg block of iron at 85.0 oC is dropped into the container.
Find the final equilibrium temperature of the system, assuming no heat loss or gain to the
surroundings. [specific heat (J/kg.K): copper = 390, iron = 470]
A)
B)
C)
D)
E)
27.2 oC
33.1 oC
52.5 oC
13.7 oC
95.3 oC
Ans:
∆𝑄+ = ∆𝑄−
(𝑚𝑐𝑢 × 𝑐𝑐𝑢 + 𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 )�𝑇𝑓 − 20� = 𝑚𝑖𝑟𝑜𝑛 × 𝑐𝑖𝑟𝑜𝑛 × �85 − 𝑇𝑓 �
𝑇𝑓 =
Q4.
𝑇𝑓 =
85 × 𝑚𝑖𝑟𝑜𝑛 × 𝑐𝑖𝑟𝑜𝑛 + (𝑚𝑐𝑢 × 𝑐𝑐𝑢 + 𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 ) × 20
𝑚𝑐𝑢 × 𝑐𝑐𝑢 + 𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 + 𝑚𝑖𝑟𝑜𝑛 × 𝑐𝑖𝑟𝑜𝑛
85 × 0.25 × 470 + (0.5 × 390 + 0.18 × 4190) × 20
= 27.16℃ = 27.2℃
0.5 × 390 + 0.18 × 4190 + 0.25 × 470
An ideal diatomic gas is initially at 27 oC and a pressure of 1.0 atm. It is compressed
adiabatically to one fifth of its initial volume. What is the final pressure (in atm) of the
gas?
A)
B)
C)
D)
E)
9.5
2.5
5. 0
2.9
15
Ans:
𝛾
𝑃𝑖 𝑉𝑖
=
𝛾
𝑃𝑓 𝑉𝑓
𝛾
𝑉𝑖
5 1.4
⇒ 𝑃𝑓 = 𝑃𝑖 � � = 1 � � = 9.5 𝑎𝑡𝑚
𝑉𝑓
1
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Q5.
Zero Version
Page: 3
A block of ice of mass 1.00 kg at 0 oC is heated until it becomes water at 100 oC.
Calculate the entropy change of the ice until it becomes water at 100 oC.
A)
B)
C)
D)
E)
Ans:
∆𝑆 =
=
Q6.
Final-133
Wednesday, August 13, 2014
2.53 kJ/K
1.57 kJ/K
1.22 kJ/K
2.31 kJ/K
1.47 kJ/K
𝑚𝑖𝑐𝑒 × 𝐿𝑓
373
+ 𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 × 𝑙𝑛 �
�
273
273
373
1 × 333 × 103
+ 1 × 4190 × 𝑙𝑛 �
� = 2529.4 𝐽/𝐾
273
273
∆𝑆 = 2.53 × 103 𝐽/𝐾
Three point charges are arranged as shown in FIGURE 1. Charges Q 1 and Q 3 are
positive, charge Q 2 is negative, and the magnitudes of Q 1 and Q 2 are equal. What is the
direction of the net electrostatic force on Q 3 due to Q 1 and Q 2 ?
A)
B)
C)
D)
E)
Negative y direction
Positive y direction
Positive x direction
Negative x direction
The net force on Q3 is zero
Figure 1
Ans:
𝐴
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 4
Q7.
Two point charges, 2.50 µC and – 3.50 µC, are placed on the x axis, as shown in
FIGURE 2, one at the origin and the other at x = 0.600 m, respectively. Find the
coordinate of a point on the x axis where the electric field due to these two charges is
zero.
A)
B)
C)
D)
E)
Ans:
– 3.27 m
+ 3.27 m
+ 0.280 m
+ 0.880 m
– 0.880 m
Figure 2
𝐸1
P 𝐹2
d
𝐴𝑡 𝑃 |𝐸1 | = |𝐸2 |
|𝑞2 |
𝑘𝑞1
𝑘𝑞2
0.6 + 𝑑
3.5
�|𝑞1 |
�|𝑞2 |
�
�
� 2�= �
�
⇒
=
⇒
=
=
= 1.18
(0.6 + 𝑑)2
|𝑞1 |
𝑑
𝑑
0.6 + 𝑑
𝑑
2.5
0.6 + 𝑑
0.6
= 1.18 ⟹ (1.18 − 1)𝑑 = 0.6 ⇒ 𝑑 =
= 3.27 𝑚 = −3.27 𝑚
𝑑
0.18
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Q8.
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 5
A small sphere with a mass of 4.00 × 10-6 kg and carrying a charge of 50.0 nC hangs
from a string near a very large, charged insulating sheet, as shown in FIGURE 3. The
charge density on the surface of the sheet is 2.50 nC/m2. What is the angle (θ) which the
string makes with the vertical?
A)
B)
C)
D)
E)
10.2o
19.8o
70.2o
79.8o
45.2o
Tsinθ
Figure 3
Tcosθ
Tsinθ T
Ans:
𝑇𝑠𝑖𝑛𝜃 = 𝑞𝐸
𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔
𝜎
𝑞 × 2𝜀
𝑞𝐸
0
=
𝑇𝑎𝑛𝜃 =
𝑚𝑔
𝑚𝑔
qE
mg
σ
q × 2ε
50 × 10−9 × 2.50 × 10−9
0
θ = tan−1 �
� = tan−1 �
� = 10.2°
mg
4 × 10−6 × 9.8 × 2 × 8.85 × 10−12
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 6
Q9.
Two point charges, carrying equal magnitude of charge, are arranged in three
configurations as shown in FIGURE 4. Rank the configurations according to the work
done by an external agent to separate the charges to infinity, greatest first.
Figure 4
A) 1, 2, 3
B) 3, 1, 2
C) 3, 2, 1
D) 1, 3, 2
E) 2, 3, 1
Ans:
𝑊𝑒𝑥𝑡 = −𝑈𝑖 = −
Q10.
𝑊1−𝑒𝑥𝑡
𝑘𝑞1 𝑞2
𝑑
𝑘𝑞 2
𝑘𝑞 2
𝑘𝑞 2
=+
; 𝑊2−𝑒𝑥𝑡 = +
; 𝑊3−𝑒𝑥𝑡 = −
𝑑
2𝑑
2𝑑
The electric potential at points in a xy plane is given by V = 1.5 x2 − 3.5 y2 (where V is
in volts and x and y are in meters). In unit-vector notation, what is the electric field (in
V/m) at the point (2.0, 3.0) m?
A) −6.0 iˆ + 21 ˆj
B) +6.0 iˆ − 21 ˆj
C) −6.0 iˆ − 21 ˆj
D) +6.0 iˆ + 21 ˆj
E) −21iˆ + 6.0 ˆj
Ans:
𝐸�⃗ (2,3) = 𝐸𝑥 𝚤⃗ + 𝐸𝑦 𝚥⃗
𝐸𝑥 = −
𝜕𝑉
𝜕𝑉
= −3𝑥; 𝐸𝑦 = −
= +7𝑦
𝜕𝑥
𝜕𝑦
𝐸𝑥 (𝑥 = 2) = −6 ; 𝐸𝑦 (𝑦 = 3) = +21
𝐸�⃗ = 𝐸𝑥 𝚤⃗ + 𝐸𝑦 𝚥⃗ = −6𝚤⃗ + 21𝚥⃗
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 7
Q11.
In FIGURE 5, potential difference across 5.0 µF capacitor is V ab = 28 V. What is the
charge on the 11 µF capacitor?
A)
B)
C)
D)
E)
Ans:
77 µC
140 µC
63 µC
54 µC
86 µC
Figure 5
𝑞5𝜇𝐹 = 𝐶5𝜇𝐹 × 𝑉 = 5 × 10−6 × 28 = 140𝜇𝐹
𝐶9−11 = 𝐶9𝜇𝐹 + 𝐶11𝜇𝐹 = (9 + 11)𝜇𝐹 = 20𝜇𝐹
𝑉11 =
Q12.
𝑞5𝜇𝐹
140 × 10−6
=
= 7𝑉
𝐶9−11
20 × 10−6
𝑞11 = 𝐶11 × 𝑉11 = 11 × 10−6 × 7 = 77𝜇𝐶
What is the magnitude of the electric field in a copper wire, of radius 1.02 mm, that is
needed to cause a 2.75 A current to flow? Copper has a resistivity of 1.72 × 10-8 Ω.m.
A)
B)
C)
D)
E)
Ans:
0.0145 V/m
0.317 V/m
0.0464 V/m
0.0547 V/m
0.108 V/m
𝜌
𝑉 𝑖𝑅
𝜌
𝑖×𝜌
2.75 × 1.72 × 10−8
𝑅
= ;𝐸= =
= 𝑖� � =
=
= 0.0145 𝑉/𝑚
𝜋 × (1.020 × 10−3 )2
𝐴
𝑙
𝑙
𝐴
𝜋𝑟 2
𝑙
King Fahd University of Petroleum and Minerals
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c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Q13.
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 8
Find the current through 4.00 Ω resistor in the circuit, shown in FIGURE 6 (Note that
the emfs are ideal).
Figure 6
A)
B)
C)
D)
E)
1.11 A
5.21 A
6.32 A
0.75 A
2.67 A
𝑖1
𝑖2
Ans:
20 − 2𝑖1 − 14 + 4𝑖3 = 0
𝑖1 = 3 + 2𝑖3 → (1)
36 − 5𝑖2 − 4𝑖3 = 0
5𝑖2 = 36 − 4𝑖3
𝑖2 = 7.2 − 0.8𝑖3 → (2)
𝑖2 = 𝑖1 + 𝑖3
𝐹𝑟𝑜𝑚 (2) 7.2 − 0.8𝑖3 = 𝑖2 = 𝑖1 + 𝑖3 ⇒ 7.2 − 0.8𝑖3 = 3 + 2𝑖3 + 𝑖3 = 3 + 3𝑖3
3𝑖 + 0.8𝑖3 = 7.2 − 3 = 4.2
𝑖3 =
Q14.
4.2
= 1.11 𝐴
3.8
Five resistors are connected as shown in FIGURE 7. What is the potential difference
V A −V B , if the current through the 2.70 Ω resistor is 1.22 A?
A)
B)
C)
D)
E)
15.0 V
19.0 V
10.7 V
22.2 V
5.54 V
𝑖1 Figure 7
Ans:
𝑉𝐴 − 𝑉𝐵 = (𝑖1 + 1.22)𝑅𝑒𝑞
𝑖1 =
𝑅𝑒𝑞
2.7 × 1.22
= 0.51 𝐴
6.5
2.70 × 6.50
= 3.20 +
+ 3.60 = 8.71 Ω
2.70 + 6.50
1.22 𝐴
𝑉𝐴 − 𝑉𝐵 = (𝑖1 + 1.22)𝑅𝑒𝑞 = (0.51 + 1.22) × 8.71 = 15.0 𝑉
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 9
Q15.
In FIGURE 8, four identical resistors are connected to an ideal battery. S1 and S2 are
switches. Which of the following actions would result in the minimum power dissipated
in resistor A?
Figure 8
A)
B)
C)
D)
E)
Ans:
𝑃𝑚𝑖𝑛 =
keeping both switches open
closing S1 only
closing both switches
closing S2 only
The answer depends on the value of the
emf of the battery
𝑉2
𝑅𝑚𝑎𝑥
𝑅𝑚𝑎𝑥 𝑓𝑜𝑟 𝑏𝑜𝑡ℎ 𝑠𝑤𝑖𝑡ℎ𝑐𝑒𝑠 𝑜𝑝𝑒𝑛 𝑜𝑛𝑙𝑦 !
𝑅𝑚𝑎𝑥 = 𝑅 + 𝑅 = 2𝑅
Q16.
Each of the two real batteries in FIGURE 9 has an emf of 20.0 V and an internal
resistance r = 0.500 Ω. What is the potential difference across each battery if R=
5.00 Ω and the current in resistor R is 3.80 A?
Figure 9
1.9 A
A) 19.1 V
B) 10.1 V
C) 5.50 V
1.9 A
D) 9.55 V
E) 20.0 V
Ans:
𝑉 = 𝜀 − 𝑖𝑟
= 20 − 1.9 × 0.5 = 20 − 0.95 = 19.05 = 19.1 𝑉
King Fahd University of Petroleum and Minerals
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c-20-n-30-s-0-e-1-fg-1-fo-0
i = 3.8 A
Phys102
Coordinator: A.A.Naqvi
Q17.
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 10
A 200 µF capacitor, that is initially uncharged is connected in series with a 6.00
Ω k
resistor and an emf source with ε = 200 V and negligible resistance. The circuit is closed
at t = 0. What is the rate at which electrical energy is being dissipated in the resistor at t =
1.00 s?
A)
B)
C)
D)
E)
1.26 W
2.22 W
0.500 W
3.42 W
5.11 W
Ans:
𝑃 = 𝑖 2 (𝑡) 𝑅 ; 𝑖(𝑡) =
𝑖(𝑡 = 1𝑠) =
𝜀 −𝑡/𝑅𝐶
𝑒
𝑅
200 −1/�6000×200×10−6 �
1 (−1/1.2)
1 −0.833
𝑒
=
𝑒
=
𝑒
= 0.0145 𝐴
6000
30
30
30
𝑃 = 𝑖 2 (𝑡) 𝑅 = (0.0145)2 × 6000 = 1.26 𝑊
Q18.
An electron has a velocity of 6.0 × 106 m/s in the positive x direction at a point where the
magnetic field has the components B x = 3.0 T, B y = 1.5 T, and B z = 2.0 T. What is the
magnitude of the acceleration of the electron at this point? (Ignore gravity)
A)
B)
C)
D)
E)
2.6 × 1018 m/s2
1.6 × 1018 m/s2
2.1 × 1018 m/s2
3.2 × 1018 m/s2
3.7 × 1018 m/s2
Ans:
�⃗ ��
�6.0 × 106 𝚤⃗ × �3𝚤⃗ + 1.5𝚥⃗ + 2𝑘
𝐹𝐵
𝑞𝑒 (𝑣 × 𝐵)
−19
𝑎⃗ =
=
= −1.6 × 10
𝑚𝑒
𝑚𝑒
9.1 × 10−31
=
−1.6 × 10−19 × 6 × 106
�⃗ �� = −1.05 × 1018 �1.5𝑘
�⃗ − 2𝚥⃗�
�1.5(𝚤⃗ × 𝚥⃗) + 2�𝚤⃗ × 𝑘
9.1 × 10−31
|𝑎⃗| = 1.05 × 1018 × √6.25 = 2.6 × 1018 𝑚/𝑠 2
King Fahd University of Petroleum and Minerals
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c-20-n-30-s-0-e-1-fg-1-fo-0
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Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 11
Q19.
In FIGURE 10, current i = 50.0 mA is set up in a loop having two radial lengths and two
semicircles of radii a =10.0 cm and b =20.0 cm with a common center P. What is the
magnitude of the loop’s magnetic dipole moment?
Figure 10
A)
B)
C)
D)
E)
-3
3.93×10
1.16×10-2
5.11×10-3
4.22×10-3
2.32×10-3
2
A.m
A.m2
A.m2
A.m2
A.m2
Ans:
𝜇 = 𝑁𝑖𝐴
(𝑟𝑎2 + 𝑟𝑏2 )
𝜇 = 𝑖𝜋
2
(0.12 + 0.22 )
= 50 × 10−3 𝜋 2
2
= 3.93 × 10−3 𝐴 ∙ 𝑚2
Q20.
A particle (m = 3.3 × 10 − 27 kg, q = 1.6 × 10 − 19 C) is accelerated from rest through a 10
kV potential difference and then moves perpendicular to a uniform magnetic field with
magnitude B = 1.2 T. What is the radius of the resulting circular path?
A)
B)
C)
D)
E)
17 mm
19 mm
20 mm
10 mm
9.0 mm
Ans:
𝐹𝐵 = 𝑞𝑣𝐵 = 𝐹𝑅 =
𝑅=
=
𝑚𝑣 2
𝑚𝑣
1
2𝑞𝑉
⇒𝑅=
𝑏𝑢𝑡 𝑞𝑉 = 𝑚𝑣 2 ⇒ 𝑣 = �
𝑅
𝑞𝐵
2
𝑚
𝑚𝑣
𝑚
2𝑞𝑉
=
�
𝑞𝐵 𝑞𝐵
𝑚
1 2𝑚𝑉
1 2 × 3.3 × 10−27 × 10000
�
�
=
𝐵
𝑞
1.2
1.6 × 10−19
𝑅 = 0.017 𝑚 = 17 𝑚𝑚
King Fahd University of Petroleum and Minerals
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Final-133
Wednesday, August 13, 2014
Zero Version
Page: 12
Q21.
A 5.0 A current is flowing in a 5.0 m long wire which is aligned along the direction of the
unit vector 0.60 î + 0.80 ĵ . The wire is placed in a region where the magnetic field is

given by B = 0.80 k̂ (T). The magnitude of the magnetic force on the wire is:
A)
B)
C)
D)
E)
Ans:
20 N
23 N
59 N
17 N
40 N
𝐹𝐵
�⃗
�⃗ � ⇒ 𝐹𝐵 = 𝑙𝑖�𝑙⃗ × 𝐵
�⃗ � = 5 × 5 × (0.6𝚤⃗ + 0.8𝚥⃗) × 0.8𝑘
= 𝑖 �𝑙⃗ × 𝐵
𝑙
�⃗ � + 0.8 × 0.8�𝚥⃗ × 𝑘
�⃗ �� = 25�0.48(−𝚥⃗) + 0.64(𝚤⃗)�
𝐹⃗𝐵 = 25 �0.6 × 0.8�𝚤⃗ × 𝑘
= 25 × 0.64 𝚤⃗ − 25 × 0.48 𝚥⃗ = 16𝚤⃗ − 12𝚥⃗
|𝐹𝐵 | = �162 + 122 = 20 𝑁
Q22.

An electron moving with a velocity v = 5.0 × 10 m/s î enters a region of space where

perpendicular electric and magnetic fields are present. The electric field is E = −
4
1.0×10 ĵ . What magnetic field will allow the electron to pass through the region,
undeflected?
7

A) B =
− ( 2.0 × 10−4 T ) kˆ

B) B =
+ ( 2.0 × 10−4 T ) ˆj

C) B =
− ( 2.0 × 10−4 T ) ˆi

D) B =
+ ( 2.0 ×10−4 T ) kˆ

E) B =
+ ( 5.0 ×10−4 T ) kˆ
Ans:
𝐴
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 13
Q23.
The loop shown in FIGURE 11 consists of two semicircles 1 and 2 (with the same
center) and two radial lengths. The semicircle 1 lies in the xy plane and has a radius of
10.0 cm, and the semicircle 2 lies in the xz plane and has a radius of 4.00 cm. If the
current i in the loop is 0.500 A, what is magnitude of the magnetic field at point P,
located at the center of the loops?
Figure 11
A)
B)
C)
D)
E)
Ans:
4.23
3.71
1.37
2.92
6.45
µT
µT
µT
µT
µT
�⃗
𝐵𝑛𝑒𝑡 = 𝐵2 𝚥⃗ + 𝐵1 𝑘
𝐵1 =
𝜇0𝑖
4𝜋 × 10−7 × 0.5
=
= 15.71 × 10−7 𝑇
4𝑅1
4 × 0.1
𝐵2 =
𝜇0𝑖
4𝜋 × 10−7 × 0.5
=
= 39.27 × 10−7 𝑇
4𝑅1
4 × 0.04
|𝐵𝑛𝑒𝑡 | = �𝐵12 + 𝐵22 = �(15.71)2 + (39.27)2 × 10−7 𝑇 = 4.23𝜇𝑇
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 14
Q24.
FIGURE 12 shows three long, parallel, current-carrying wires carrying equal magnitude
of current. The directions of currents I1 and I3 are out of page. The arrow labeled F
represents the magnetic force per unit length acting on current I3 due to the other two
wires and is given by F = (− 0.220 iˆ − 0.220 ĵ ) (N/m) . What are the magnitude and
direction of the current I2, if the distance d = 1.00 mm?
A)
B)
C)
D)
E)
33.2 A, into the page
33.2 A, out of the page
22.1 A, into the page
22.1A, out of the page
11.4 A, out of the page
Ans:
𝐹 = −0.22𝚤⃗ − 0.22𝚥⃗ = 𝐹13 𝚤⃗ − 𝐹23 𝚥⃗
2
𝜇0 𝑖 2 4𝜋 × 10−7 𝑖 2
|𝐹23 | = 0.22 =
=
×
2𝜋𝑑
2𝜋
𝑑
0.22 × 𝑑
0.22 × 10−3
�
𝑖=�
=
= 33.2 A into the page
2 × 10−7
2 × 10−7
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Figure 12
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 15
Q25.
Two infinitely long current-carrying wires are parallel to each other, as shown in
FIGURE 13. The magnetic field at point P, which is midway between them, is 0.12 mT
out of the page. If the current I 1 = 12 A, what is magnitude and direction of I 2 ?
A)
B)
C)
D)
E)
Ans:
Figure 13
21 A, to the right
21 A, to the left
3.0 A, to the right
3.0 A, to the left
15 A, to the left
𝐵𝑛𝑒𝑡 = −0.12 × 10−3 𝑇 = 𝐵1 + 𝐵2
𝐵2 = −0.12 × 10−3 − 𝐵1
2
𝜇0 𝑖1
4𝜋 × 10−7 × 12
𝐵1 =
=
= 16 × 10−5 𝑇
2𝜋𝑑
2𝜋 × 0.015
𝐵2 = 0.12 × 10−3 + 0.16 × 10−3 = 0.28 × 10−3
𝜇0 𝑖 2
2𝜋𝑑𝐵2
2𝜋𝑑𝐵2
⇒ 𝑖2 =
=
2𝜋𝑑
𝜇0
4𝜋 × 10−7
2
𝑑𝐵2
0.015 × 0.28 × 10−3
𝑖2 =
=
= 21𝐴
2 × 10−7
2 × 10−7
𝐵12 =
Q26.
Which of the solenoids described below has the greatest magnetic field along its axis?
A)
B)
C)
D)
E)
a solenoid of length 2L, with 2N turns and a current 2I
a solenoid of length L, with N turns and a current I
a solenoid of length L, with N/2 turns and a current I
a solenoid of length L/2, with N/2 turns and a current I
a solenoid of length 2L, with N turns and a current I/2
Ans:
𝐴
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 16
Q27.
A cylindrical conductor of radius R = 2.50 cm carries a 2.50 A current along its length.
This current is uniformly distributed throughout the cross sectional area of the conductor.
Find the distances from the center of the wire (inside and outside the wire) at which the
value of magnitude of magnetic field equals half of its maximum value.
A)
B)
C)
D)
E)
1.25 cm, 5.00 cm
1.50 cm, 6.50 cm
1.00 cm, 4.00 cm
1.25 cm, 7.50 cm
1.50 cm, 3.00 cm
Ans:
𝐵𝑜𝑢𝑡 =
𝜇0𝑖
1 𝜇0 𝑖
1
1
1
= �
� = 𝐵𝑚𝑎𝑥 ⇒
=
2𝜋𝑅𝑐𝑜𝑚 2 2𝜋𝑅
2
𝑅𝑐𝑜𝑚
2𝑅
𝑅𝑜𝑢𝑡 = 2𝑅 = 2 × 2.50 = 5.0 𝑐𝑚
𝜇0𝑖
𝐵𝑚𝑎𝑥
1 𝜇0𝑖
𝑟 1
𝑅 2.50
�
𝑟
=
=
×
⇒
=
⟹
𝑟
=
=
= 1.25 𝑐𝑚
2𝜋𝑅 2 𝑅
2
2 2𝜋𝑅 𝑅 2
2
2
𝐵𝑖𝑛 = �
𝐵𝑖𝑛 at r = 1.25 cm, 𝐵𝑜𝑢𝑡 at r = 5.00 cm
Q28.
A 20-turn circular coil of radius 5.00 cm is placed in a magnetic field perpendicular to the
plane of the coil. The magnitude of the magnetic field varies with time as B(t) = 0.0100 t
+ 0.0400 t2, where B is in Tesla and t is in second. Calculate the magnitude of the
induced emf in the coil at t = 5.00 s.
A)
B)
C)
D)
E)
Ans:
64.4 mV
40.4 mV
32.2 mV
23.5 mV
30.5 mV
𝑑𝐵
𝑑𝐵
(𝑡 = 5𝑠) = 0.01 + 0.4 = 0.41
= 0.01 + 0.08𝑡;
𝑑𝑡
𝑑𝑡
|𝜀| = 𝑁
𝑑𝐵
𝐴 = 20 × 0.41 × 𝜋(0.05)2 = 0.0644 V
𝑑𝑡
|𝜀| = 64.4 𝑚𝑉
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: A.A.Naqvi
Final-133
Wednesday, August 13, 2014
Zero Version
Page: 17
Q29.
A conducting bar of length L moves to the right on two frictionless rails, as shown in
FIGURE 14. A uniform magnetic field, directed into the page, has a magnitude of 0.25
T. Assume R = 10 Ω, L = 0.45 m. At what constant speed should the bar move to produce
a 9.0 mA current in the resistor?
Figure 14
A) 0.80 m/s
B) 0.85 m/s
C) 0.75 m/s
D) 0.70 m/s
E) 0.65 m/s
Ans:
|𝜀| = 𝐵𝐿𝑣 = 𝑖𝑅
𝑣=
𝑖𝑅 9 × 10−3 × 10
=
= 0.8 𝑚/𝑠
𝐵𝐿
0.25 × 0.45
Q30.
Two rectangular loops of wire lie in the same plane as shown in FIGURE 15. If the
current I in the outer loop is counterclockwise and increases with time, which of the
following statements is CORRECT about the current induced in the inner loop?
A)
B)
C)
D)
It is clockwise.
It is counterclockwise.
It is zero.
Its direction depends on the dimensions of
the loop.
E) Its direction is fluctuating with time.
Ans:
𝐴
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Figure 15