physics 111 spring 2006 exam # 1
1/8 PHYSICS 1308 SPRING 2008 EXAM # 3 Thursday, April 10, 2008
Solutions
CONCEPTUAL QUESTIONS 1. A long wire carries a current I as shown. What is the direction of the current in the circular conducting loop when I is: (a) Increasing (b) Decreasing I
(a) When the current is increasing, the field inside the ring (which points into the page ) also increases, thus the field induced should point out of the page. By the righthand rule the current in the loop must be CCW. (b) A similar reasoning shows that if the current in the wire decreases the induced current in the loop is CW
2. Can an induced electric field exist in the absence of a conductor? Explain your answer. Yes, a particular example is an electromagnetic wave. The changing magnetic field induces an electric field that also changes in time and space. Since the electromagnetic wave can propagate in empty space, we have an induced electric field without a conductor.
3. If in an electromagnetic wave the electric field is parallel to the xaxis and the magnetic field is parallel to the z axis, what is the direction of propagation of the wave ? (Justify your answer) The direction of propagation of an electromagnetic wave is given by c = E× B . Thus the wave propagates parallel to: i×−k = j which means that the wave propagates parallel to the positive yaxis.
2/8
4. Explain Maxwell's contribution to Ampere's Law. Maxwell added the displacement current to Ampere's Law. This current just states that a changing (in time) electric field induces a magnetic field.
5. Light passes from a material with index of refraction n 1 into one with index of refraction n 2 with n 1 n 2 . Compared with the incident ray, what happens to the refracted ray?
(a) It bends toward the normal (b) It is undeflected (c) It bends away from the normal Explain your selection. n 1 n 2 means by the definition of index of refraction that v 1 v 2 , i.e. light propagates from a
region of lower velocity to a region of higher velocity. As we saw in class, this means that the refracted way bends away from the normal.
NUMERICAL QUESTIONS An automobile has a vertical radio antenna 1.20 m long. The automobile travels at 65 km/h on a horizontal road where the Earth's magnetic field is 50 T directed toward the north. 6. The direction that the automobile should move to generate maximum motional emf in the antenna is : A. North to South
B. East to West
C. West of North
D. North of East
In this case, in order to have a maximum emf we need to maximize the flux, and this is done if the direction of motion is perpendicular to the field. Therefore, we need to choose East to West. 7. The magnitude of this emf is _____mV A. 1
B. 2
C. 4
D.6
E. 8
3/8
A. 3.2
B. 2.5
C. 1.0
x x
x
x
x x
x
x
x x
x x
x x
x
x
x
x x x
F
x
l
x
R
x x
x
8. A conducting rod of length l moves on two horizontal, frictionless rails as shown in the figure. A constant force of 1.00 N moves the bar at 2.00 m/s through a magnetic field B that is directed into the page. The current in in the 8.00 resistor is _____A
x
dt
=− B l v =−50 ×10 −6 T ×1.20m ×18m / s =−1.08 ×10−3 V
x
−d B
x
=
D. 0.5
E. 0.1
Since the rod is moving with constant velocity, the power delivered by the change is the magnetic flux is equal to the power put in by the motion of the rod. Note that this is just consequence of conservation of energy. Hence : Pgenerated = P emf F⋅v = I 2 R Fv
R
=
1.00N × 2.00m / s 8.00
= 0.5 A
9. The current in a 90.0 mH inductor changes with time as I =1.00 t 2 −6.00 t in (SI units). The the induced emf at t = 4.00 s is ____mV A. 720
=
−d B dt
=−L
B. 720 dI dt
=−90×10 H −3
C. 0 d dt
2
D. 180
E. 180
1.00 t −6.00t =−90×10 H 2.00t −6.00 −3
now, we just need to evaluate this function at the required time: =−90×10−3 H 2.00 4.0 s−6.00=−180×10 −3 V
___________________________________________________________________________
4/8 10. A 22.0 F capacitor is connected to an AC source with V rms=120 V , f = 60.0 Hz. The maximum current going through the capacitor is _____A. B. 1.40
A. 0.52 I max =
V max XC
C.2.20
D. 3.45
E. 5.52
=V max C= 2 V rms C= 2 120V 2 60Hz22×10 F=1.4 A −6
11. The resistance of a light bulb that uses an average power of 75 W when connected to a 60.0 Hz power source having a maximum voltage of 170 V is ____ A. 385 2
Pave = I rms V rms=
C. 192
B. 220 V rms R 2
120V 75W
V / 2 =
D. 153
E. 28
2
max
R
=192
12. In a purely inductive AC circuit, L = 25.0 mH and the rms voltage is 150.0 V. The rms current in the circuit is _____A ( Assume that the frequency is 60.0 Hz) A. 100 I rms =
V rms XL
=
V rms L
B. 85
=
C. 50
150.0V 2 60Hz 25.0×10 H −3
=15.9 A
D. 34
E. 16
5/8 The RC circuit in the figure below has a resistance R = 90.0 and a capacitance C = 8.00 nF. The AC source has a frequency of 60 Hz and a maximum voltage of 170 V. Refer to the circuit below for questions 13 and 14. C
R
V
V out
13. The phase between V out and the voltage source is _____rad . Use the voltage source as reference. B. 1.57
A. 1.57
C. 0
D. 90
E. 90
First notice that V out is the same as the voltage across the resistor. This voltage is always in phase with the current going through it. This current is the same as the current flowing across the capacitor which always leads the voltage source by 90º . Thus V out is ahead 90º from the the voltage source. ( 1.57 rad = 90º ) 14. The average power delivered by the AC source is _____W A. 0
B. 5
C. 7
D. 12
E. 20
The power delivered by the AC source is equal to the power consumed in the resistor. The 2 average power is given by the equation Pave = I rms R . So we need first to find I rms . This current can be found using the current of the RLC circuit with the inductive reactance equal to zero. Thus: I rms =
V rms Z
=
V rms 2
R X
= 2 C
V rms
2
1
2
R
C
6/8 We are ready: 2 V rms R
Pave =
R 2
1
2
120 V 90.0
= 2
90 2
C
2
1 2 60Hz 8.0×10−9 F
=11.77×10−6 W
15. A parallelplate capacitor has square plates 10 cm on a side and 0.50 cm apart. It the voltage across the plates is increasing at the rate of 220 V/ms, the displacement current in the capacitor is −12 2 2 ___A. 0 =9×10 C / N⋅m A. 4.0 I d =0
d E dt
=0 A
B. 7.0 dE dt
=0 A
C. 10.0
D. 13.0
E. 16.0
1 dV 1 = 9×10−12 C 2 / N⋅m 2 0.10 m 2 220V d dt ms
1ms
−3
10 s
1 0.0050m
I d =3.96×10 −6 A
16. A laser produces an average power of 7.0 W in a light beam of 1.0 mm in diameter. The maximum electric field of the laser light is ______kV/m ( c =3 ×10 8 m / s and 0 =4 ×10 −7 T⋅m / A )
A. 82
B. 96
C. 105 P
D. 152
E. 263
E 2max
I= = A 2 0 c
2 P0 c A
=E max =
2⋅7W⋅4 ×10−7 Tm / A⋅3 ×108 m / s −3
2
0.5×10 m
=82 ×10 3 V / m
17. A microwave oven operates at 2.4 GHz. The distance between wave crests is _____m A. 0.0625
B. 0.125
C. 0.625
D. 0.725
E. 1
7/8 Recognizing that the distance between wave crests is equal to the the wavelength we can write: c= f 8 3× 10 m / s / 2.4 ×10 9 Hz = 0.125 m =
18. Light propagating in air strikes a transparent crystal at incidence angle of 35º. If the angle of refraction is 22º, the speed of light inside the crystal is _____ ×108 m/s A.
3.00
B. 2.95
C. 2.47
D. 2.12
E. 1.96
Assuming that light propagates in air at the same velocity that in vacuum we have by Snell's Law sin 350 = n2 sin 22 0 therefore: c
sin 22 0
sin 35
0
8
= v 2 =1.96×10 m/s
19. A ray of light is incident of the surface of a block of clear ice at an angle of 40º with the normal. Part of the light is reflected and part is refracted. The angle between the reflected and refracted light is _____º. ( n ice =1.309 )
A. 0.0
B. 29.4
40º
C. 40.0
D. 75.2
E. 110.6
By Snell's law: 40º
sin 40 0 =1.309 sin 0 = 29.41
therefore : = 180º – 40º – 29.4º = 110.6º
8/8 _______________________________________________________________________________ 20. A plastic light pipe has an index of refraction of 1.53. For total internal reflection the minimum angle of incidence if the pipe is in water is ____º. ( n water =1.333 ) A. 0.0 sin C =
B. 30.5
C. 54.2
D. 60.6
E. 90.0
n2 n1
C =60.60
0
Write your answers here: 6. B
7.A
8.D
9.E
10.B
11.C
12. E
13.B
14.D
15.A
16.A
17.B
18.E
19.E
20.D
Solutions
CONCEPTUAL QUESTIONS 1. A long wire carries a current I as shown. What is the direction of the current in the circular conducting loop when I is: (a) Increasing (b) Decreasing I
(a) When the current is increasing, the field inside the ring (which points into the page ) also increases, thus the field induced should point out of the page. By the righthand rule the current in the loop must be CCW. (b) A similar reasoning shows that if the current in the wire decreases the induced current in the loop is CW
2. Can an induced electric field exist in the absence of a conductor? Explain your answer. Yes, a particular example is an electromagnetic wave. The changing magnetic field induces an electric field that also changes in time and space. Since the electromagnetic wave can propagate in empty space, we have an induced electric field without a conductor.
3. If in an electromagnetic wave the electric field is parallel to the xaxis and the magnetic field is parallel to the z axis, what is the direction of propagation of the wave ? (Justify your answer) The direction of propagation of an electromagnetic wave is given by c = E× B . Thus the wave propagates parallel to: i×−k = j which means that the wave propagates parallel to the positive yaxis.
2/8
4. Explain Maxwell's contribution to Ampere's Law. Maxwell added the displacement current to Ampere's Law. This current just states that a changing (in time) electric field induces a magnetic field.
5. Light passes from a material with index of refraction n 1 into one with index of refraction n 2 with n 1 n 2 . Compared with the incident ray, what happens to the refracted ray?
(a) It bends toward the normal (b) It is undeflected (c) It bends away from the normal Explain your selection. n 1 n 2 means by the definition of index of refraction that v 1 v 2 , i.e. light propagates from a
region of lower velocity to a region of higher velocity. As we saw in class, this means that the refracted way bends away from the normal.
NUMERICAL QUESTIONS An automobile has a vertical radio antenna 1.20 m long. The automobile travels at 65 km/h on a horizontal road where the Earth's magnetic field is 50 T directed toward the north. 6. The direction that the automobile should move to generate maximum motional emf in the antenna is : A. North to South
B. East to West
C. West of North
D. North of East
In this case, in order to have a maximum emf we need to maximize the flux, and this is done if the direction of motion is perpendicular to the field. Therefore, we need to choose East to West. 7. The magnitude of this emf is _____mV A. 1
B. 2
C. 4
D.6
E. 8
3/8
A. 3.2
B. 2.5
C. 1.0
x x
x
x
x x
x
x
x x
x x
x x
x
x
x
x x x
F
x
l
x
R
x x
x
8. A conducting rod of length l moves on two horizontal, frictionless rails as shown in the figure. A constant force of 1.00 N moves the bar at 2.00 m/s through a magnetic field B that is directed into the page. The current in in the 8.00 resistor is _____A
x
dt
=− B l v =−50 ×10 −6 T ×1.20m ×18m / s =−1.08 ×10−3 V
x
−d B
x
=
D. 0.5
E. 0.1
Since the rod is moving with constant velocity, the power delivered by the change is the magnetic flux is equal to the power put in by the motion of the rod. Note that this is just consequence of conservation of energy. Hence : Pgenerated = P emf F⋅v = I 2 R Fv
R
=
1.00N × 2.00m / s 8.00
= 0.5 A
9. The current in a 90.0 mH inductor changes with time as I =1.00 t 2 −6.00 t in (SI units). The the induced emf at t = 4.00 s is ____mV A. 720
=
−d B dt
=−L
B. 720 dI dt
=−90×10 H −3
C. 0 d dt
2
D. 180
E. 180
1.00 t −6.00t =−90×10 H 2.00t −6.00 −3
now, we just need to evaluate this function at the required time: =−90×10−3 H 2.00 4.0 s−6.00=−180×10 −3 V
___________________________________________________________________________
4/8 10. A 22.0 F capacitor is connected to an AC source with V rms=120 V , f = 60.0 Hz. The maximum current going through the capacitor is _____A. B. 1.40
A. 0.52 I max =
V max XC
C.2.20
D. 3.45
E. 5.52
=V max C= 2 V rms C= 2 120V 2 60Hz22×10 F=1.4 A −6
11. The resistance of a light bulb that uses an average power of 75 W when connected to a 60.0 Hz power source having a maximum voltage of 170 V is ____ A. 385 2
Pave = I rms V rms=
C. 192
B. 220 V rms R 2
120V 75W
V / 2 =
D. 153
E. 28
2
max
R
=192
12. In a purely inductive AC circuit, L = 25.0 mH and the rms voltage is 150.0 V. The rms current in the circuit is _____A ( Assume that the frequency is 60.0 Hz) A. 100 I rms =
V rms XL
=
V rms L
B. 85
=
C. 50
150.0V 2 60Hz 25.0×10 H −3
=15.9 A
D. 34
E. 16
5/8 The RC circuit in the figure below has a resistance R = 90.0 and a capacitance C = 8.00 nF. The AC source has a frequency of 60 Hz and a maximum voltage of 170 V. Refer to the circuit below for questions 13 and 14. C
R
V
V out
13. The phase between V out and the voltage source is _____rad . Use the voltage source as reference. B. 1.57
A. 1.57
C. 0
D. 90
E. 90
First notice that V out is the same as the voltage across the resistor. This voltage is always in phase with the current going through it. This current is the same as the current flowing across the capacitor which always leads the voltage source by 90º . Thus V out is ahead 90º from the the voltage source. ( 1.57 rad = 90º ) 14. The average power delivered by the AC source is _____W A. 0
B. 5
C. 7
D. 12
E. 20
The power delivered by the AC source is equal to the power consumed in the resistor. The 2 average power is given by the equation Pave = I rms R . So we need first to find I rms . This current can be found using the current of the RLC circuit with the inductive reactance equal to zero. Thus: I rms =
V rms Z
=
V rms 2
R X
= 2 C
V rms
2
1
2
R
C
6/8 We are ready: 2 V rms R
Pave =
R 2
1
2
120 V 90.0
= 2
90 2
C
2
1 2 60Hz 8.0×10−9 F
=11.77×10−6 W
15. A parallelplate capacitor has square plates 10 cm on a side and 0.50 cm apart. It the voltage across the plates is increasing at the rate of 220 V/ms, the displacement current in the capacitor is −12 2 2 ___A. 0 =9×10 C / N⋅m A. 4.0 I d =0
d E dt
=0 A
B. 7.0 dE dt
=0 A
C. 10.0
D. 13.0
E. 16.0
1 dV 1 = 9×10−12 C 2 / N⋅m 2 0.10 m 2 220V d dt ms
1ms
−3
10 s
1 0.0050m
I d =3.96×10 −6 A
16. A laser produces an average power of 7.0 W in a light beam of 1.0 mm in diameter. The maximum electric field of the laser light is ______kV/m ( c =3 ×10 8 m / s and 0 =4 ×10 −7 T⋅m / A )
A. 82
B. 96
C. 105 P
D. 152
E. 263
E 2max
I= = A 2 0 c
2 P0 c A
=E max =
2⋅7W⋅4 ×10−7 Tm / A⋅3 ×108 m / s −3
2
0.5×10 m
=82 ×10 3 V / m
17. A microwave oven operates at 2.4 GHz. The distance between wave crests is _____m A. 0.0625
B. 0.125
C. 0.625
D. 0.725
E. 1
7/8 Recognizing that the distance between wave crests is equal to the the wavelength we can write: c= f 8 3× 10 m / s / 2.4 ×10 9 Hz = 0.125 m =
18. Light propagating in air strikes a transparent crystal at incidence angle of 35º. If the angle of refraction is 22º, the speed of light inside the crystal is _____ ×108 m/s A.
3.00
B. 2.95
C. 2.47
D. 2.12
E. 1.96
Assuming that light propagates in air at the same velocity that in vacuum we have by Snell's Law sin 350 = n2 sin 22 0 therefore: c
sin 22 0
sin 35
0
8
= v 2 =1.96×10 m/s
19. A ray of light is incident of the surface of a block of clear ice at an angle of 40º with the normal. Part of the light is reflected and part is refracted. The angle between the reflected and refracted light is _____º. ( n ice =1.309 )
A. 0.0
B. 29.4
40º
C. 40.0
D. 75.2
E. 110.6
By Snell's law: 40º
sin 40 0 =1.309 sin 0 = 29.41
therefore : = 180º – 40º – 29.4º = 110.6º
8/8 _______________________________________________________________________________ 20. A plastic light pipe has an index of refraction of 1.53. For total internal reflection the minimum angle of incidence if the pipe is in water is ____º. ( n water =1.333 ) A. 0.0 sin C =
B. 30.5
C. 54.2
D. 60.6
E. 90.0
n2 n1
C =60.60
0
Write your answers here: 6. B
7.A
8.D
9.E
10.B
11.C
12. E
13.B
14.D
15.A
16.A
17.B
18.E
19.E
20.D
