Download 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