Download Here is the solution of Exam 1.

Ryoichi Kawai BLAZER ID kawai
Write your name also in the back of the last page.
NAME
PH202-U4 Test 1 (June. 28, 2012, 3:00PM-5:00PM)
• You may not open the textbook nor notebook.
• A letter size information sheet may be used.
• A calculator may be used. However, mathematics or physics formula programmed in a calculator may
not be used.
• Solve seven problems out of ten. Clearly cross out three problems you do not solve. If less than
three problems are crossed out, the first seven problems that are not crossed out are graded even if the
answer is empty.
• Write down, reasoning, calculation and answer in the blank space after each problem. Use the backside of the sheet if necessary.
• Answers without reasonable explanation or convincing mathematical derivation will receive no point
even if your answers coincide with the correct ones. Substantial partial credit may be given to correct
reasoning and mathematical procedures even when your final answers are wrong. However, if your
final answers are too obviously wrong, a partial credit may not be given.
Important Physical Constants:
elementary charge: e = 1.60 × 10−19 C
electron’s mass: me = 9.11 × 10−31 kg
proton’s mass: mp = 1.67 × 10−27 kg
1
permittivity of free space: 0 = 4πk
= 8.85 × 10−12 C 2 /(N · m2 ) where k = 8.99 × 109 N · m2 /C 2
permeability of free space: µ0 = 4π × 10−7 T · m/A
acceleration due to gravity at the surface of the earth: g = 9.80 m/s2
speed of flight: c = 3.00 × 108 m/s
1
2
3
4
7
8
9
10
5
6
Total
1
y
1. A charge Q = −2.0 µC is placed at the coordinate origin as
shown in Figure.
F
A
y=0.30 m
(a) Draw an arrow indicating the direction of electric field at
point A.
E
(b) Find the magnitude and direction of the electric field.
(c) Another charge q = −5.0 µC is placed at point A. Find the
magnitude and direction of the force exerted on q.
x=−0.40 m
Q
x
(a) Consider a test charge at point A. Since the test charge is positive, it is attracted to the negative charge
Q. Hence, the direction of the electric field is from A to Q. (See Figure).
p
(b) The distance from Q to A = r = (−0.40)2 + (0.30)2 = 0.50 m. Then, the magnitude of the electric
field is
E=
k|Q|
(8.99 × 109 ) × (2.0 × 10−6 )
=
= 7.2 × 104 N/C
2
r
(0.50)2
Since the direction is on the diagonal line as shown in drawing, we find
0.30
= −36.9◦
arctan
−0.40
The direction is 36.9◦ below the +x axis.
(c) The magnitude of the force is
F = |q|E = (5.0 × 10−6 ) × (7.2 × 104 ) = 0.36 N
The direction of the force is opposite to that of the electric field. Hence, 36.9◦ above the −x axis (or 143.1◦
from the +x axis.)
2
2. The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a
small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge
density on the inner and outer walls has a magnitude of 7.1 × 10−6 C/m2 .
(a) What is the magnitude of the electric field within the cell membrane?
(b) Find the magnitude of the electric force that would be exerted on a potassium ion (K+ ) placed inside
the membrane.
(a)
E=
Q
7.1 × 10−6
=
= 8.0 × 105 N/C
ε0 A
8.85 × ×10−12
(b)
F = eE = (1.6 × 10−19 ) × (8.0 × 105 ) = 1.3 × 10−13 N
3
3. Two protons are initially very far apart and moving in the opposite direction with the same speed,
v = 4.0 × 104 m/s. They collide head-on. What is the distance of closest approach?
1
ke2
ke2
Initial Energy = 2 × mv 2 = mv 2 and final energy =
. Since the energy conserves,
= mv 2 .
2
r
r
Solving this equation for r, we find
r=
(8.99 × 109 ) × (1.60 × 10−19 )2
ke2
=
= 8.6 × 10−11 m
2
mv
1.67 × 10−27
4
4. The interior of a biological cell is separated from the exterior by a membrane. The membrane is
1 × 10−8 m thick and its dielectric constant is 5. The outer surface of the membrane is charged positively,
and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor.
(a) Assuming that the membrane acts like a parallel plate capacitor with a plate area of 6 × 10−6 m2 ,
what is its capacitance?
(b) Suppose that the electric potential outside the cell is higher than that inside the cell by 0.07 V . How
much work is done by the electric force when a sodium ion (Na+ ) moves from the outside to the
inside?
(a)
C=
kε0 A
5 × (8.85 × 10−12 ) × (6 × 10−6 )
=
= 2.7 × 10−8 F
d
1 × 10−8
(b) Since W = EP Ei − EP Ef we find the work
W = eV = (1.6 × 10−19 ) × 0.07 = 1.1 × 10−20 J
You can also find the same result from force. To find the force, first we need to find the electric field.
E=
0.07
V
=
= 7.0 × 106 N/C
d
1 × 10−8
F = eE = (1.60 × 10−19 ) × (7.0 × 106 ) = 1.1 × 10−12 N
Then, the work is
W = F d = (1.1 × 10−12 ) × (1 × 10−8 ) = 1.1 × 10−20 J
5
5. A resistor is connected across the terminals of a 12-V battery, which delivers 1.5 × 105 J of energy to
the resistor in six hours. What is the resistance of the resistor?
Work done during a time period t is given by W = P t. The power P is determined by P =
W =
V 2t
. Solving this equation for R, we obtain
R
R=
(12)2 × (6 × 3600)
V 2t
=
= 21 Ω
W
1.5 × 105
6
V2
and thus
R
6. Two different circuits are constructed using the same components; three
resistors, R1 = 60 Ω, R2 = 30 Ω, and R3 = 20 Ω, and a 24-V battery as
illustrated in Figures.
(a) Find the electric current though R2 in circuit (a).
(b) Find the electric current though R2 in circuit (b).
(c) Find the power dissipated by the circuit (a).
(d) Find the power dissipated by the circuit (b).
(a) Since the three resistors are connected in series, the equivalent resistance is
R123 = R1 + R2 + R3 = 110 V . Then, the current drawn from the battery is
I=
V
24
=
= 0.22 A
R123
110
For series connection, this current flows through all resistors. Hence, the current in R2 is 0.22 A.
(b) Since the three resistors share the same voltage, they are in parallel. Then, the current in R2 is simply
given by
I=
24
V
=
= 0.80 A
R2
30
(c) Using the equivalent resistance,
P =
V2
(24)2
=
= 5.2 W
R123
110
V2
. Hence, the total power is
Ri
V2 V2 V2
1
1
1
2
P =
+
+
=V
+
+
= (24)2 (1/60 + 1/30 + 1/20) = 58 W
R1
R2
R3
R1 R2 R3
(d) Each resistor disspates power of
7
7. A positively charged particle of mass m = 6.0 × 10−8 kg is travelling due east with a speed of
v = 90 m/s and enters a uniform magnetic field whose magnitude is B = 0.30 T . The particle moves
through one-quarter of a circle in a time of t = 2.0 × 10−3 s, at which time it leaves the field heading due
south. All during the motion the particle moves perpendicular to the magnetic field.
(a) What is the magnitude of the magnetic force acting on the particle?
(b) Determine the magnitude of its charge.
mv 2
(a) Since the particle makes a uniform circular motion, there must be centripetal force Fc =
. This
r
magnitude of the force is provided by magnetic force FB = qBv. Hence, Fc = FB We cannot evaluate
these equation because the value of neither r nor q is given. Using given information, we must find them.
πr
The velocity and time gives are the length of arc s = vt. Since it is a quarter of circle, s =
. Hence,
2
πr
2vt
= vt. Solving this equation for r, we obtain r =
. Plug in r into the centripetal force,
2
π
FB = Fc =
mv 2 π
mvπ
(6.0 × 10−8 ) × 90π
=
=
= 4.2 × 10−3 N
2vt
2t
2 × (2.0 × 10−3 )
(b) Since FB = Fc ,
q=
Fc
4.2 × 10−3
=
= 1.6 × 10−4 C
vB
90 × 0.30
8
8. Figure shows a rectangular N = 20-turn coil of wire (ABCD). Its
dimensions is BC = 0.10 m by AB = 0.05 m. It is mounted in the xy
plane and hinged along one long side (AD) on the y axis. It carries a current
of i = 0.20 A. The direction of the current is shown in Figure. A uniform
magnetic field of magnitude B = 0.50 T , is applied at 30◦ from the x axis
as shown in Figure.
A
B
D
C
(a) Find the direction and magnitude of the force exerted on the wire AB. The direction should be
expressed with the axis names. For example, if the direction is to the left, it should be expressed as
−x direction.
(b) Which edges (AB, BC, CD, DA) contribute to the torque around the hinge. List all of them.
(c) Find the magnitude of the torque. You must derive it from basic physics principles.
(d) In which direction does the coil rotate, viewed from the top, clockwise or counterclockwise?
(a) Both current and magnetic field are on the horizontal plane (xz plane). Hence the force must be
perpenticular to the xz plane. Using Right-Hand rule, it is −y direction.
(b) The force on AB and BC are parallel to the hinge. Hence, they do not contribute to the torque. DA is
the hinge and thus the force on it does not contribute the torque (because the lever arm = 0). Only the force
on CD contributes the torque.
(c)
τ = N ABI sin φ = 20 × (0.10 × 0.05 m2 ) × (0.50 T ) × (0.20 A) × sin(60◦ ) = 8.7 × 10−3 N · m
(d) The current is vertical downward and the magnetic field is on the xz plane. Hence, the force is also on
the xz and perpendicular to the magnetic field. Using Right Hand rule, the force is in the direction that the
coil rotates in the clockwise direction.
9
⃗
B( into page)
9. A loop of wire has the shape shown in the drawing. The top part of the
wire is bent into a semicircle of radius r = 0.20 m. The normal to the plane
of the loop is parallel to a constant magnetic field (φ = 0◦ ) of magnitude
B = 0.50 T . The bottom wire includes a resistor of R = 4.0 Ω as shown in
Figure. Starting with the position shown in Figure, the semicircle is rotated
through half a revolution in a time of t = 0.20 s.
r
R
(a) What is the change ∆Φ in the magnetic flux that passes through the
loop?
(b) What is the average current in the resistor?
(c) In which direction, the current flows in the resistor, right or left?
R
(a) The area inside the loop is reduced by a circle of radius r. Hence, the change of the magnetic flux is
∆Φ = BAf cos(0) − BAi cos(0) = B(Af − Ai ) = −B(πr2 ) = −0.50 × π × (0.20)2 = −0.063 W b
(b) Using the Faraday’s law, E = −
I=
∆Φ
. From the Ohm’s law, we fin the current
∆t
∆Φ
0.063
E
=−
=
= 0.079 A
R
∆tR
0.2 × 4.0
(c) Since the magnetic flux decreases, the Lenz’s law tells that the current flows in such a way that the
magnetic flux increases. That means the current induces the magnetic field into page inside the loop.
Hence, the current flows to the left inside the resistor.
10
10. Two circuits contain an emf produced by a moving metal rod, like that
shown in Figure. The speed of the rod is the same in each circuit, but the
bulb in circuit 1 has one-half the resistance of the bulb in circuit 2. The
circuits are otherwise identical.
I
Conducting rail
L
⃗v
I
(a) Find the ratio E1 /E2 of the emfs.
(b) Find the ratio I1 /I2 of the currents in the circuits.
(c) If the speed of the rod in circuit 1 were twice that in circuit 2, what would be the ratio P1 /P2 of the
powers in the circuits?
(a) EMF is give by E = BLv. Since the two circutes share the same vales for B, L, and v, the two circuits
E1
have the same emf. Hnece,
= 1.
E2
(b)
I1
E1 /R1
E1 R2
=
=
·
=1·2=2
I2
E2 /R2
E2 R1
(c) Now, the two circuits have different speed. Then, the ratio of EMF is no longer 1.
E1
BLv1
v1
=
=
=2
E2
BLv2
v2
P1
I1 E1
I1 E1
E1 R2 E1
=
=
·
=
·
=
P2
I2 E2
I2 E2
E2 R1 E2
E1
E2
2
·
11
R2
= (2)2 · 2 = 8
R1