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Worksheet 10
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1. Consider the circuit shown in Figure.
(a) Assuming that currents, I1 , I2 , and I3 flaw from right to left
through the registers as shown in Figure. Write down the
Kirchhoff’s junction rule using I1 , I2 , and I3 .
(b) Write down the voltage relation for the upper loop using the
Kirchhoff’s loop rule.
(c) Write down the voltage relation for the lower loop using the
Kirchhoff’s loop rule.
(d) Find the magnitude of current in the 2.0-Ω resistor.
(e) Find the direction of actual current in the 2.0-Ω resistor (Answer must be one of ”To the right”, ”To the left”, or ”No
current”.)
See Figure for the potential drop across the resistors.
(a)
I1 = I2 + I3
(1)
3.0I1 + 2.0I2 = 3.0
(2)
(b)
(c)
1.0 + 4.0I3 = 2.0I2 + 4.0
→
2.0I2 − 4.0I3 = −3.0
(3)
5.0I2 + 3.0I3 = 3.0
(4)
(d) Substituting EQ (1) into EQ (2),
3.0(I2 + I3 ) + 2.0I2 = 3.0
→
Now we eliminate I3 by calculating 4 × EQ (4) + 3 × EQ(3). Then, we have
26.0I2 = 3.0
→
I2 =
1
3.0
= 0.12 A
26.0
2. A charged particle of mass m = 2.0 × 10−8 kg and charge q =
~ perpendicular
+8.0 µC is traveling in a uniform magnetic field B
to the page. The initial velocity is 40 m/s in the +y direction.
(See Figure.) Particle detectors a and b are placed as shown
in Figure. The distances between the particle and detectors are
d1 =d2 =0.20 m.
(i) Due to the presence of the magnetic field, the particle
reaches to detector b. Find the magnitude and direction
(“into the page” or “out of the page”) of the magnetic field.
~ perpendicular to the magnetic field
(ii) When an electric field E
~
B is applied, the particle travels in a straight line to detector
a. Find the magnitude and direction (+x, -x, +y, or -y)
of the electric field. [Assume that the particle is initially
located at the same place with the same velocity as part
(i) of this problem. Note that the magnetic field is still in
tact.]
(a)
The particle travels in a circle of radius R = d2 /2 = 0.10m to the detector b as shown in Figure. The
magnetic field provides centripetal force. Hence,
qvB =
mv 2
R
→
B=
mv
(2.0 × 10−8 kg)(40m/s)
=
= 1.0 T
qR
(8.0 × 10−6 C)(0.10m)
(b) Initially, the direction of the magnetic force exerted on the particle is +x. In order for the particle
move in a straight line to the detector a, the magnetic force must be cancelled by electric force. Hence, the
direction of the electric field is −x and its magnitude is
qE = qvB
→
EvB = (40m/s)(1.0T ) = 40 N/C
2
3. A rock is located at the bottom of a 2.0-m deep pond 1.2 m from an edge of the pond as shown in
Figure. The pond is filled with water. A boy whose eye is 1.5-m above the ground sees the rock just at the
edge. A fish at the bottom of the pond sees the image of the rock made by reflection at the surface. (Index
of refraction for water is 1.333.)
(a) Draw a ray from the rock to the boy’s eye.
(b) How far is the boy from the edge of the pond?
(c) Draw a ray from the rock to the fish.
(d) At least how far is the fish from the rock?
(a) See rays shown in Figure.
(b) First, we need to find θ1 (See Figure).
tan θ1 =
1.2m
= 0.60
2.0m
θ1 = tan−1 (0.60) = 31.0◦
→
Using the Snell’s law, we find θ2 as follow:
sin θ2 = nwater sin θ1 = 1.333 sin 31.0◦ = 0.686
→
θ2 = sin−1 (0.686) = 43.3◦
Now, we calculate the distance x in Figure,
x = h tan θ2 = (1.5m) tan 43.3c irc = 1.4 m
(c) First, we calculate the critical angle for the interface between air and water.
sin θc =
nair
1
=
= 0.75
nwater
1.333
→
θc = sin−1 (0.75) = 48.6◦
The distance between the fish and the rock is
2y = 2(2.0m) tan 48.6◦ = 4.5 m
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