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Trang 1SOLUTIONS, KEYS AND SCORES For Questions in Final Exam of Principles of Physics 2
Edited by: Phan Gia Anh Vu
Date of Exam: July 24th 2020
In electric field, a charge of q will be exerted by the force F q E
In this case, q is negative, so F
is parallel but in opposite direction with E
0.5
2 C A = D
In the conducting spherical shell, the inner and the outer surface will possess
negative and positive charge, respectively
Applying the Gauss’s law (
0
in
q
) for the Gaussian spheres that contain the points A, B, C and D, it can be found that:A BB0;C0
0.5
Applying the relationship between E and B:
B
E
c
1.5010 7 3.0010845.0N/C
E
0.5
The positions of minima are given by y Ltan
The angles in which the minima occur can be found from: dsinmWith
small angles tan sin
The position of the first dark fringe is:
d L
10 5
10 5 0
6
y L
0.5
5 The magnetic field create by the current in the wire
is given by
a
I B
2
0
At the position of the loop, the magnetic field is perpendicular to and out of
the page (as in the figure on the right) The farer a
point from the wire is, the weaker the magnetic
field is
a) If the loop moves away from the wire, the
magnetic flux through the loop will decrease So the induced current in the
loop is counterclockwise
b) If the loop moves to the wire, the magnetic flux through the loop will
increase So the induced current in the loop is clockwise
0.5
0.5
B
Trang 26 Applying the right hand rule, we can find the
direction of the magnetic field created by the
current I1 at the position of I2 as in the figure
The force exerting on the wire carrying I2 is
vertical and upwards
As the result, the two wires attract each other
0.5
0.5
7 Let number the charges from 1 to 4 as in the figure The electric fields created by the
charges q1, q2, q3, q4 are E 1, E 2, E 3, E 4
, respectively
We can see that E 2 E 4; E 1 2 E 3
So the total electric field at the square center is:
3 4 3 2
E
The magnitude of E 3
is:
2
6 9
2
2
2 05 0
10 6 10 9
2 2
a
kq
So the magnitude ofE
is: E E 3 3 1 30 108Nm2/C2
/C N.m ˆ ˆ 10 19
b) If a fifth charge is located at O , it will be exerted by the force:
6 00 10 6 9 19 107 i ˆ ˆ j 551 i ˆ ˆ j N
E q
(the red arrow
in the above figure)
The magnitude of this force is: F 780 N
0,5
0,5
0,5
0,5
8
a) The current I1 creates a magnetic field B 1
that is perpendicular to the page and directs into the page
In order to have the zero total magnetic field at O,
the current I2 must create the magnetic field B 2
so that: B 2 B 1
Thus, the current I2 has to be in opposite direction
of the current I1 and have the magnitude that:
0
2 1
1
r
I r
I
or
1 1 2
2
r
I r
I
12
9
1 1
2
r
r
So I2 is counterclockwise and has the magnitude of: 3.75 A
b) If the direction of I2 remains unchanged but its magnitude is doubled, then the total magnetic field at O is nonzero Now, B 1 B2, so the total field has the same direction with B 2
(out of the page) The magnitude of B
is:
T 10 62 2 09
0
75 3 2 12 0
5 2
10 4 2
5 7
2 2 1
1 0 2
r
I r
I I
0,5
0,5
0,5
0,5
9 a) From the condition of constructively reflection on the thin film:
m
2 2
×
2 3
1 4
O
Trang 3It can be found that:
m
nt
2
The first three longest wavelengths are corresponding with the three smallest value of m: 1, 2 and 3
3
10 00 5 38 1 2
μm 690 0 2
10 00 5 38 1 2
μm 38 1 m 10 38 1 1
10 00 5 38 1 2
7 3
7 2
4 7
1
b) From these wavelengths, there are 2 wavelengths in the visible spectrum
μm 460 0 μm;
690 0
0,5
0,5
0,5