Upload
epic-win
View
218
Download
0
Embed Size (px)
Citation preview
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 1/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
Physics B: FyBNVC08G-level Complementary Test on Electricity and Magnetism Ch 16-19, & 20
Instructions:
The Test Warning! There are more than one version of the test.
At the end of each problem a maximum point which one may get for a correct solution
of the problem is given.
Tools Approved formula sheets, ruler, and graphic calculator. You may use your green
personalized summary-formula-booklet which has your name on it. This should be
submitted along with the test.
Time: 13:10-16:00
Solutions to the problems must be answered on the original test paper.
Limits: This tests gives at maximum 22 points. It is only a complementary test to the G-testgiven earlier this week. The passing scores or qualification for a higher grade is judged
individually based on the student’s performance in the other two tests.
Please answer in this paper and as clear as possible. Do not forget the UNITS!Enjoy it! Behzad
Name:
16-20 E C C C, V C RC C, Q B, I B, Q Q, E Q, V
P 1 2 3 4a 4b 4c 4d 5 6 7a 7b Sum
G 1 1 1 1 2 2 2 2 2 2 2 18
G
16-20 B, Q B, Q Ch 16-20 Grade
P 8a 8a Total
G 2 2 22
G
Information regarding G-Test:
Teacher’s Comments
Student’s Comments
© [email protected] ☺Free to use for educational purposes. Not for sale. 1/7
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 2/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale. 2/7
1. In the Bohr’s model of the hydrogen atom, an electron circles a proton at a distance
m1110− . Find the electric field that acts on the electron. Show your calculations in
the space available below. [1/0]
1. In the Bohr’s model of the hydrogen atom, an electron circles a proton at a distance
m1110− . Find the electric field that acts on the electron. Show your calculations in
the space available below. [1/0]
3.5 ×3.5 ×
Solutions Answer: m
V
E 11
101.5 ×=
( ) m
V
r
Qk E
1122199
22
199
211
199
2101.510513.0
1009.28
106.1109
103.5
106.1109 ×=×=
×
×⋅×=
×
×⋅×== +−
−
−
−
−
2. Ten identical capacitors of capacitance F 0. are connected in series. Find the
equivalent capacitance of the combination. Show your calculations in the space
available below. [1/0]
20
Solutions Answer: F eqC μ 0.2=
F F
C C
C C C C C C C
eqeq
eqeq
μ 0.210
.20
10
101...
1111
321
==⇔=⇔=⇔+++=
3. We triple the distance between the plates of a parallel plate capacitor of capacitance C .
Find the capacitance of the new capacitor. Show your calculations in the space available
below.
Answer: 32
C
C =
32
1
3002
C
d
A
d
AC =⋅== ε ε
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 3/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale. 3/7
4. A parallel plate capacitor, a resistor and a battery are connected as illustrated in the
figure below. The two parallel plates of the capacitor are
4. A parallel plate capacitor, a resistor and a battery are connected as illustrated in the
figure below. The two parallel plates of the capacitor are mm500.0 apart. The plates
are connected through a Ωk .100 resistor to a battery. When the plates are fully
charged, the electric field between the plates is mkV / ..200
a) Find the electric potential difference between the plates. Show your calculations in
the space available below. [1/0]
Suggested solution: Answer: kV 100.0V =
Data: mmmd 4100.550.0 −×== , mV mkV E /10.200/.200
3×==
d E V ⋅= ⇔ kV V d E V 100.0.1001000.510.200 43 ==×××=⋅= −
b) If the parallel capacitor has an area of 200 and a plate separation of .4 cm
mm500.0 . Find its capacitance . [2/0]
Suggested Solutions: Answer: pF C 08.7=
( ) pF F d
AC 08.71008.7
105
1041085.8 12
4
412
0 =×=
×
×⋅×== −
−
−−ε [2]
c) Find the time constant of the circuit, and explain its significance. Show your
calculations in the space available below. [2/0]
Suggested solution: Answer: ns.708=τ
Data: Ω×=Ω= 310.100.100 k R F 121008.7 −×=; C
RC =τ ⇔ ( )( ) nssF .70810.7081008.710.100 9123 =×=×Ω×= −−τ
ns.708=τ [2]
d) Find total charge on the positive plate of the capacitor, when the capacitor is fullycharged. Show your calculations in the space available below. [2/0]
Suggested solution:Answer: nC 708.0Q =
Data: ; pF F C 08.71008.7 12 =×= −V kV 10010.0V ==
V C Q ⋅= ⇔ ( )( ) nC C F V Q 708.01008.71008.7100 1012 =×=×= −−
nC C Q 708.01008.7 10 =×= − [1]
R
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 4/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale. 4/7
5. A straight wire carrying a5. A straight wire carrying a A0.25 current has a length cm0.20 between the poles of a
magnet at an angle °.53 as illustrated in the figure
below. The magnetic field is fairly uniform at T 00.4 .
The magnitude and direction of the force on the wire.
Find the magnitude and the direction of the magnetic
force on the wire. Show your calculations in the spaceavailable below. [2/0]
Suggested solution: Answer: N F 0.16=
Data: A I 0.25= ,
mcm 200.00.20 ==l , °= .53θ , T B 00.4=
Problem: ?=F
B I F r
lrr
×= ⇔ θ sin B I F l= ⇔ N F 0.16127sin00.4200.00.25 =×××= [1/0]
Answer: The total force on the wire is N F 0.16= out of the page of
the paper towards the reader using the right hand rule. [1/0]
6. A charged particle moves in a circular path of radius r in a magnetic field B . The
velocity of the charged particle is tripled. Compare the radius of the circular path of the
charged particle. Explain. Show your calculations in the space available below. [2/0]
Suggested solution: Answer: BQ
vmr
⋅⋅=
BQ
vmr
r
vm BvQ
r
vmF
BvQF
⋅⋅=⇔⋅=⋅/⋅⇔
⎪⎩
⎪⎨
⎧
⋅=
⋅⋅= /2
2
Answer: According to BQ
vmr
⋅⋅= , if the velocity of the charge particle is
tripled, radius of its orbit must also be tripled. [2/0]
A0.25
T 00.4
°.53
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 5/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale. 5/7
7. A charge C 7. A charge C 0. charge is placed cm0.16 from an identical C 50+ μ 0.50+ charge.
a) What is the electric field at the point midway between the charges? [2/0]
b) What is the electric potential at the point midway between the charges? [2/0]
Suggested solution:Data:
C QQQ 0.5021 +=≡=C Q 0.103 +=
a) The electric field at the midway between the charges is zero, thisis due to the fact that the electric field due to C Q 0.501 += at the
midway is to the right, and that of C Q 0.502 += is to the left. The
magnitude of the electric field due to each charge midwaybetween the charges is identical. But 1 E
ris to the right, while 2 E
ris
to the left. Therefore, the vector sum of them must be zero, i.e.
021 =+= E E E rrr
. [2/0]
b) The electric potential at the midway is:r
Qk
r
Qk
r
Qk V M 221 =+=
mcmcm
r 080.00.82
0.16===
V V M
66
9 10125.1
08.0
105´1092 ×=
×⋅××=
−
[2/0]
C Q 0.501 +=C Q μ 0.502 +=
cm0.16
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 6/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale. 6/7
8. An electron is shot into the region between the poles of a magnet. The initial velocity of
the electron is
8. An electron is shot into the region between the poles of a magnet. The initial velocity of
the electron is sm /1000.3 6×
perpendicular to the direction of the
magnetic field. The magnetic field is
uniform and its magnitude is T 500.0 as
illustrated in the figure.
Data: C eQ 191060.1 −×−=−= ,
kgme
311011.9 −×=
a) Find the direction and magnitude of the magnetic force on the electron.
Explain. Show your calculations in the space available below.
[2/0]
Suggested solution: Answer: N F 131040.2 −×=
Data: C eQ 191060.1 −×−=−= , sm /1000.3 6×=v , T B 500.0=
The electron will experience a magnetic force which is into thepage of the paper and away from the reader of these words.Therefore, the electron will be deflected into the page and awayfrom the reader. The magnitude of the force is:
BvQF rrr
×⋅= ⇔
N BvQF 13619 1040.2500.01000.31060.190sin −− ×=××××=⋅⋅⋅=
b) Describe the path of the electron in the uniform magnetic field, and find thenecessary information regarding the path. [2/0]
Suggested solution: Answer: The electron will experience acentripetal acceleration, and as a result it will rotate in a circular
path of radius m BQ
vmr μ 2.34=
⋅
⋅= , normal to the page of the
paper
Data: C Qe
191060.1−×−= , smv /1000.3 6×= , T B 500.0= ,
kgme
311011.9 −×=
BvQF
rrr
×⋅= ⇔
BQ
vmr vmr BQ
r
vm BvQ
r
vmF
BvQF
⋅
⋅=⇔⋅=⋅⋅⇔⋅=⋅⋅⇔
⎪⎩
⎪⎨
⎧
⋅=
°⋅⋅⋅=2
2
90sin
mm BQ
vmr μ 2.341016.34
500.01060.1
1000.31011.9 6
19
631
=×=××
×××=
⋅
⋅= −
−
−
The electron will experience a centripetal acceleration, and as a
result it will rotate in a circular path of radius m BQ
vmr μ 2.34=
⋅
⋅= ,
normal to the page of the paper
C eQ 19106.1 −×−=−= smv /100.3 6⋅=
T B 500.0=
8/3/2019 Solutions+G Level+Complementary++FyBCh16 19,+20NVCO08+Electricity+and+Magnetism+Published
http://slidepdf.com/reader/full/solutionsg-levelcomplementaryfybch16-1920nvco08electricityandmagnetismpublished 7/7
Solution G-level Complementary Test FyBCh16-19-20NVC08 Electricity & Magnetic Field NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale. 7/7