NATIONAL UNIVERSITY OF SINGAPORE DEPARTMENT OF PHYSICS ADVANCED PLACEMENT TEST (SAMPLE)

Comments

Transcription

NATIONAL UNIVERSITY OF SINGAPORE DEPARTMENT OF PHYSICS ADVANCED PLACEMENT TEST (SAMPLE)
NATIONAL UNIVERSITY OF SINGAPORE
DEPARTMENT OF PHYSICS
ADVANCED PLACEMENT TEST
(SAMPLE)
PC1143
PHYSICS III
MMM-YYYY----Time Allowed: 2 Hours
INSTRUCTIONS TO CANDIDATES
1. This paper contains 5 short questions in Part I and 3 long questions in Part II. It
comprises 9 printed pages.
2. The total marks for Part I is 40 and that for Part II is 60.
3. This is a CLOSED BOOK test.
4. Non-Programmable calculator is allowed to be used.
5. Answer ALL the questions.
6. Answers to the questions are to be written in the answer booklets.
7. A table of constants and mathematical formulae is attached.
2
PART I
This part of the examination paper contains five short-answer questions on pages 2 to 4.
Answer ALL questions.
Question 1
Consider the box-like Gaussian surface G as shown in Figure 1.
y
y2
x2
x1
z1
x
z1
z2
z2
z
x1
x2
Figure 1
The bottom face is in the xz-plane, while the top face is in the horizontal plane passing
through y2  2.00 m . Suppose x1  1.00 m , x2  4.00 m , z1  1.00 m , and z2  3.00 m ,
and G lies in an electric field given by

E  [(6.00 x  10.0)iˆ  3.00 ˆj  zkˆ] N C ,
with x and z in metres and  is a constant. If G encloses a net charge of  48.00 C , find  .
[8]
Question 2

(a) Consider a particle P of mass m and positive charge q moving with velocity v in a

uniform static magnetic field B .

(i) Describe the magnitude and direction of the magnetic force that B exerts on P. [3]

(ii) Can B do work on P? Explain briefly.
[2]
6
(b) A source injects an electron of speed v  1.5 10 m s into a uniform magnetic field of
magnitude B  1.0 102 T . The velocity of the electron makes an angle   30 with the
direction of the magnetic field. Find the distance d from the point of injection at which
the electron next crosses the field line that passes through the injection point.
[3]
3
Question 3
Figures 2 and 3 show two circuits, each with R  14.0  , C  6.20 F , L  56.0 mH , and a
battery having emf E  34.0 V .
E
R
C
R
E
L
a
a
b
b
C
L
Figure 2
Figure 3
The switch in Figure 2 is kept in position a for 2C s , while the switch in Figure 3 is kept in
position a for 2 L s . The switches are then separately thrown to their respective position b’s.
Here, C and  L are the capacitive and inductive time constants respectively. Find
(a) C and  L .
[2]
(b) the angular frequency of the resulting oscillations in each circuit.
[1]
(c) the current in each circuit 1.20 ms after the switches are being thrown to their respective
position b’s.
[5]
Question 4
A series R-L-C circuit (Figure 4) consists of a resistor R  100  , an inductor L  0.15 H ,
and a capacitor C  30 F connected in series to an 120 V (rms) emf source. The frequency
of the source is 60 Hz.
Figure 4
Calculate
(a) the current amplitude.
(b) the phase angle  .
(c) the average power loss.
[4]
[2]
[2]
4
Question 5
A plane electromagnetic wave travelling in the positive direction of an x-axis in vacuum has
electric field components Ex  E y  0 and
Ez  Emax cost  kx ,
with Emax  2.0 V m and    1015 s 1 .
(a) Find k.
[1]
(b) What is the magnetic field associated with the wave?
[3]
(c) Calculate the wave intensity.
[2]
2
(d) The wave uniformly illuminates a surface of area 2.0 m . If the surface totally absorbs
the wave, calculate the radiation pressure and the magnitude of the corresponding force
on the surface.
[2]
END OF PART I
5
PART II
This part of the examination paper contains three long questions on pages 5 to 9. Answer
ALL questions.
Question 6(a)
Positive electric charge Q is uniformly distributed along a non-conducting rod R with length
L, lying along the x-axis between x  0 and x  L , as shown in Figure 5.
z
P

z
y
L
L2
L2
x
Figure 5
Consider a point P that lies on a line, which goes through a point on the xy-plane with
coordinates x  L 2 and y  L 2 , and is parallel to the z-axis. P is at a height z above the
xy-plane. Show that the electric potential V at P is given by
 2 L2  4 z 2  L 
Q
.
V z  
ln 
2
2

4 0 L  2 L  4 z  L 
[7]
6
Question 6(b)
Now, we have four such rods as R, that are arranged into a square as shown in Figure 6.
z
P

z
y
L
L2
L2
x
Figure 6

Find the electric field E at point P. Explain clearly how you obtain your answer.
[9]
Question 6(c)
If an electron of mass m is placed at the centre of the square and is then displaced a small
distance z along the vertical dashed line as shown in Figure 6 z  L  , determine its
oscillating frequency. You may ignore gravity.
[4]
7
Question 7(a)

Consider a positive point charge q moving with velocity v as shown in Figure 7.
P
q

v
Figure 7
Describe the magnitude and direction of the magnetic field at point P due to q.
[6]
Question 7(b)


Hence, derive the infinitesimal magnetic field dB at point P due to a current element Idl as
shown in Figure 8.
[6]
P

Idl
Figure 8
Question 7(c)
Derive an expression for the magnetic field strength at distance d from the centre of a straight
wire of finite length L that carries current I. You have to provide a diagram of the straight
wire with clear labels of all the quantities you use in your derivation.
[8]
Question 8(a)
(i) A wire with resistivity  carries current iC . The current is increasing at the rate diC dt .
Show that there is a displacement current iD in the wire equal to 0 diC dt .
[4]
(ii) Consider a long, straight silver wire with resistivity   1.62 108   m . The current in
the wire is uniform and changing at the rate of 2000 A s when the current is 100 A.
What is the ratio of the magnitude of the magnetic field due to the displacement current to
that due to the current at a small distance r from the wire?
[2]
8
Question 8(b)
The square loop in Figure 9 is made of rigid conducting rods, each 3.00 m in length, with a
total series resistance of 10.0  . It is placed in a uniform 0.100 T magnetic field directed
perpendicularly into the plane of the paper.
0.100 T



b








F


a
c








d




F
Figure 9
The loop, which is hinged at each corner, is pulled until the separation between points b and d
is 3.00 m as shown in Figure 10.
9
0.100 T


b

a

F








d






c

F




Figure 10
This process takes 0.100 s to be completed, and is executed in such a way that the rate of
change in area bounded by abcd is a constant.
(i) What is the direction of the induced current in the loop? Justify your answer.
(ii) What is the power delivered to the resistors?
(iii)Using
[2]
[4]
Eind 
show that the induced emf Eind in the loop at any instant during the 0.100 s is given by
d
Eind  1.80 cos 2 ,
dt
where  is the angle that bc makes with the horizontal through c.
[8]
YY
END OF PART II
END OF PAPER
A.
Fundamental Physical Constants
Speed of light, c ≈ 2.998 × 108 m/s
Magnitude of charge of electron, e ≈ 1.602 × 10−19 C
Mass of electron, me ≈ 9.109 × 10−31 kg
Mass of proton, mp ≈ 1.673 × 10−27 kg
Permitivity of free space, ²0 ≈ 8.854 × 10−12 C2 · N−1 · m−2
Permeability of free space, µ0 = 4π × 10−7 Wb · A−1 · m−1
Acceleration due to gravity, g ≈ 9.807 m/s−2
B.
Solutions to a Quadratic Equation
ax2 + bx + c = 0,
x=
C.
−b ±
√
b2 − 4ac
.
2a
Derivatives
d n
x = nxn−1
dx
d
sin ax
dx
d
cos ax
dx
d ax
e
dx
d
ln ax
dx
d
ln f (x)
dx
d P (x)
dx Q(x)
= a cos ax
= −a sin ax
= aeax
=
a
x
1 d
f (x)
f (x) dx
·
¸
1
d
d
=
Q(x)
P
(x)
−
P
(x)
Q(x)
[Q(x)]2
dx
dx
=
1
D.
Power series
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x + ···
2!
3!
x3 x5 x7
x−
+
−
+ ···
3!
5!
7!
x2 x4 x6
1−
+
−
+ ···
2!
4!
6!
x3 2x5 17x7
x+
+
+
+ ···
3
15
315
x2 x3
1+x+
+
+ ···
2!
3!
x2 x3 x4
x−
+
−
+ ···
2
3
4
(1 + x)n = 1 + nx +
sin x =
cos x =
tan x =
ex =
ln(1 + x) =
E.
Integrals
Z
xn dx =
Z
xn+1
(n 6= −1)
n+1
dx
= ln x
x
Z
1
sin axdx = − cos ax
a
Z
1
cos axdx = sin ax
a
Z
1
eax dx = eax
a
Z
dx
x
√
= arcsin
2
2
a
a −x
Z
p
xdx
√
= a2 + x2
a2 + x2
Z
p
dx
√
= ln(x + a2 + x2 )
a2 + x2
Z
dx
x
1
= arctan
2
2
a +x
a
a
Z
dx
1
x
= 2√
a a2 + x2
(a2 + x2 )3/2
Z
xdx
1
= −√
3/2
2
2
2
(a + x )
a + x2
Z
dx
1
p
p
= ln
(x − a)2 + b2
(a − x) + (a − x)2 + b2
Z
(x − a)dx
1
= −p
3/2
2
2
[(x − a) + b ]
(x − a)2 + b2
Z
dx
x−a
= 2p
[(x − a)2 + b2 ]3/2
b (x − a)2 + b2
2

Similar documents