PHY2049 Spring 2005
Prof. Darin Acosta
Prof. Paul Avery
March 9, 2005
PHY2049, Spring 2005
Exam 2 Solutions
1. Two identical light bulbs A and B are connected in series to a constant voltage source.
A wire is then connected across B. What is the brightness of A relative to its former
brightness?
A light bulb will have a certain resistance R. The light it emits will be proportional to the
energy dissipated by this resistance: P = i 2 R . So when two identical light bulbs are
connected in series to a constant voltage V, the current passing through the circuit is
i = V / 2 R and the power dissipated in light bulb A is:
V2
V2
P1 =
R=
.
4R2
4R
Now when a wire is connected across light bulb B, that means that the resistance is
bypassed by a wire in parallel to the bulb (i.e. the bulb is shorted out by a jumper). The
resistance of bulb B becomes zero, and the current in the circuit is i = V / R . This current
is twice as high as before, but the power dissipated by light bulb A will be four times
more:
V2
P2 =
= 4 P1
R
Answer: 4
2. In the circuit shown, what is the current (in amps) flowing through the 18V battery?
Let i1 be the current through the 18V battery, i2 the current through the middle branch
with the 6V battery, and i3 the current through the rightmost branch.
By Kirchoff’s junction rule: i1 = i2 + i3 .
By Kirchoff’s voltage sum rule applied to the rightmost loop:
6V + 12Ωi2 = 24Ωi3
⇒ i3 = 0.5i2 + 0.25
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PHY2049 Spring 2005
By Kirchoff’s voltage sum rule applied to the leftmost loop:
18V − 6Ωi1 = 6V + 12Ωi2
18 − 6 ( i2 + i3 ) = 6 + 12i2
18 − 6i2 − 6 ( 0.25 + 0.5i2 ) = 6 + 12i2
18 − 1.5 − 9i2 = 6 + 12i2
10.5
= 0.5
21
⇒ i3 = 0.25 + 0.5i2 = 0.5
⇒ i2 =
⇒ i1 = i2 + i3 = 1.0
Answer: 1.0 Amps through 18V battery
3. An electron is accelerated from rest by a potential difference of 8.0 kV. It then enters a
uniform magnetic field of magnitude 1.0 T with its velocity perpendicular to the direction
of the field. Calculate the radius in meters of its path in the magnetic field.
Setting the centripetal acceleration, a = v 2 / r , equal to the acceleration due to the
magnetic force, FB / m = qvB / m , yields the following relation for the radius of curvature
for a charged particle moving in a uniform magnetic field:
mv
r=
qB
To get the velocity, one must realize that the kinetic energy of the electron after
accelerating from rest across an electric potential equals the initial potential energy:
1
EK = mv 2 = eV
2
⇒ mv = 2meV
2 ( 9.11× 10−31 kg )(1.6 × 10−19 ) ( 8000 V )
2meV
⇒r=
=
= 3.0 × 10−4 m
1.6 ×10−19
qB
Answer: 3.0 ×10−4 m
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PHY2049 Spring 2005
4. An infinitely long insulated wire carrying a current I = 20A is bent into the shape
shown (straight line plus circle of radius R=10 cm with the currents in the direction
shown). Find the magnitude of the field B at the center of the circle.
We calculate the field from both the long wire and the circle separately. From the long
wire, the field at the center of the circle, a distance R away, is:
µi
Bwire = 0 into the plane of the paper.
2π R
From the circular wire, the field is:
µi
Barc = 0 Φ where Φ = 2π
4π R
µi
Barc = 0 out of the plane of the paper
2R
Since the magnetic field from the arc is in the opposite direction of that from the long
wire, the total magnitude is the difference:
B = Barc − Bwire
µ0i (π − 1) 4π × 10−7 ( 20 )(π − 1)
=
=
= 8.6 × 10−5 T
2π R
2π ( 0.1)
Answer: 8.6 ×10−5
5. The magnetic torque exerted on a flat current-carrying loop of wire by a uniform
magnetic field B is:
Answer: maximum when the plane of the loop is parallel to B
The equation for the magnetic torque on a current loop is τ = µ × B , where µ = iA is the
magnetic dipole moment. So we see that the torque is independent of the exact shape of
the loop, just the area. The torque is a maximum when the magnetic dipole moment and
the magnetic field are perpendicular, which means that the normal to the loop plane is
perpendicular to B, or that the plane of the loop is parallel to B.
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PHY2049 Spring 2005
6. A solenoid is 3.0 cm long and has a radius of 0.50 cm. It is wrapped with 500 turns of
wire carrying a current of 2.0A. What is the magnetic field at the center of the solenoid?
One can apply Ampere’s Law to determine that for a very long solenoid, the magnetic
field inside is given by:
B = µ0 ni , where n is the number of windings per unit length and i is the current in the
wire. This formula will hold approximately for the center of a solenoid of finite length.
Thus,
 500 
−2
B = µ0 ni = ( 4π × 10−7 ) 
 ( 2 ) = 4.2 × 10 T
 0.03 
−2
Answer: 4.2 ×10 T
7. A capacitor with an initial potential difference of 100V is discharged through a resistor
when a switch between them is closed at t=0. At t=5 s, the potential difference across the
capacitor is 50V. What is the time constant of the circuit?
A capacitor in series with a resistor satisfies the following differential equation by
applying Kirchoff’s voltage sum rule:
dq q
R + =0
dt C
The solution is: q ( t ) = q0 e − t / RC
So the voltage on the capacitor is VC ( t ) =
q0 − t / RC
e
= V0 e− t / RC
C
Plugging in for what we know:
VC ( 0 ) = V0 = 100 V
VC ( t = 5 ) = V0 e −5/ RC = 50 V
⇒ ln 2 =
5
RC
⇒ τ RC = RC =
5
= 7.2 s
ln 2
Answer: 7.2 s
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PHY2049 Spring 2005
8. In the figure shown, two long straight wires with separation d carry currents i1 and
i2=2 i1 out of the page. At what point on the x-axis shown is the net magnetic field due to
the currents equal to zero?
The magnetic field from each wire will be in the same direction for x < 0 and x > d .
Only for 0 < x < d are the fields lines opposite in direction. (Use the right-hand rule to
get the direction). Thus, the net magnetic field is:
µ0
2π
i2 
 i1
 −
=0
 x d −x
⇒ i1 ( d − x ) = i2 x
⇒ i1 ( d − x ) = 2i1 x
⇒ d = 3x ⇒ x =
d
3
Answer: d/3
9. A horizontal power line carries a current of 7000A from south to north. Earth's
magnetic field, with a magnitude of 60µT, is directed toward the north with a dip angle
60° downward relative to the horizontal. Find the magnitude of the magnetic force acting
on a 100 m length of power line.
The magnetic field and the power line both point north, but the magnetic field points into
the Earth at an angle of 60°.
i
S
60°
B
The force on a current carrying wire is given by:
F = iL × B
So the force points west with magnitude:
F = iLB sin θ = ( 7000 )(100 ) ( 60 × 10−6 ) ( sin 60 ) = 36.4 N
Answer: 36.4 N
5
N
PHY2049 Spring 2005
10. In the figure shown, an ideal battery of EMF = 12.0 V is connected to a network of
resistances R1=6.0Ω, R2=12.0Ω, R3=4.0Ω, R_4=3.0Ω, and R5=5.0Ω. What is the current
through resistor R5 ?
We need to find the equivalent resistance of the circuit to find the total current supplied
−1
1 1 
by the battery. R1 and R2 are in parallel with equivalent resistance  +  = 4Ω .
 6 12 
These are in series with R3, so the top branch has resistance 8Ω. The bottom branch has
resistance 3+5=8Ω also, which means that the current evenly splits between the two
branches. The total equivalent resistance of two 8Ω resistors in parallel is 4Ω, so the
EMF 12
total current is i =
= = 3A
Req
4
So the current through R5 is half of this, or 1.5A.
Answer: 1.5A
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