PHY2049 Fall 2006
Prof. Yasu Takano
Prof. Paul Avery
Oct. 18, 2006
Exam 2 Solutions
1. (WebAssign 26.6) A current of 30 A flows through a copper wire whose radius is 1.25 mm.
Assuming the current is uniform over the wire cross section, calculate the electron drift speed in
m/s. Take the free electron density to be 7 × 1028 electrons/m3.
(1) 5.5 × 10−4
(2) 7.1 × 10−3
(3) 1.4 × 10−5
(4) 2.5 × 10−2
(5) 1.7 × 10−7
The total current I = JA is given by I = JA = ene vd A , where J is the current density, e is the
electronic charge, ne is the free electron density, vd is the drift velocity and A = πR2 is the area.
With the numbers provided, we obtain vd = 5.5 × 10−4 m/s.
2. (Homework 29.37) Use the information in the previous problem. What is the magnetic field (in
T) at a radius r = 1 mm?
(1) 3.8 × 10−3
(2) 7.1 × 10−3
(3) 6.0 × 10−3
(4) 2.5 × 10−4
(5) 1.6 × 10−2
The field B inside the wire can be calculated from Ampere’s law B ( 2π r ) = μ0ienc , where ienc is
the current enclosed at r = 1 mm. The enclosed current is proportional to the enclosed area, or
ienc = 30 (1/1.25 ) = 19.2 A. Solving for B yields B = 3.8 ×10−3 T.
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PHY2049 Fall 2006
3. (WebAssign 26.20) A battery is attached to a resistive rod as shown. The rod consists of two
sections of the same material of equal lengths but different radii. (The relative radii may be different than shown in the figure.) The resulting electric field in the top section, whose radius is 3
mm, is 2000 V/m. The electric field in the bottom section is 5000 V/m. What is the radius of the
bottom section in millimeters?
(1) 1.9
(2) 1.2
(3) 4.7
(4) 3.0
(5) 7.5
V
R
The key fact is that the total current through each section of the resistor is the same. The current
is I = V / R or I = V / R = ( EL ) / ( ρ L / A ) = EA / ρ . The equality of currents in the bottom and
top sections yields Eb Ab = Et At and using A = π r 2 we obtain rb = rt 2000 / 5000 = 1.9 mm.
4. (Chapter 27 from lecture) The light bulbs A and B in the circuit shown are identical. When the
switch S is closed, what happens to the intensity of the light bulbs?
(1) Nothing changes
(2) Bulb A gets brighter
(3) Bulb B gets brighter
(4) Bulb A gets dimmer
(5) Bulb B gets dimmer
A
12 V
S
12 V
B
12 V
Before the switch is closed we only have the two batteries on the right. These two batteries have
equal voltages (total of 24 V) and the two bulbs have the same resistance. Thus the voltage between them is half of 24 V or 12 V. When the switch is connected to the 12 V battery on the left,
the point stays at the same potential and nothing happens.
5. (From S2006 PHY2054 exam) An uncharged capacitor C is connected in series with a 200 volt
battery and a resistor R. The time constant of the circuit is 0.6 millisec. Immediately after the
connection is made, the current through the resistor is measured to be 0.8 A. What is the value
of C (in μF)?
(1) 2.4
(2) 37.5
(3) 0.16
(4) 25
(5) 250
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PHY2049 Fall 2006
At the instant the circuit is closed, the capacitor has no effect. Thus the resistance R is found
from R = V / I = 200/0.8 = 250Ω. Since the RC time is 0.008 sec, the value of C = 0.008/250 =
2.4 μF.
6. (WebAssign 27.33) In the circuit shown, E1 = 4 V, E2 = 1 V, R1 = 4Ω, R2 = 5Ω, R3 = 2Ω, and
both batteries are ideal. What is the rate (in watts) at which energy is dissipated in R2?
(1) 0.014
(2) 0.049
(3) 0.10
(4) 0.20
(5) 0.56
R1
E1
R2
R3 E2
Let i1 be the current going up through E1 and i2 be the current going up through E2. Going
around the two loops provides two equations for the voltages. These equations are
4 − 4i1 − 2 ( i1 + i2 ) = 0 and 1 − 5i2 − 2 ( i1 + i2 ) = 0 . Solving for i2 yields, i2 = 1/19 so the power dissipated by R2 is P = i22 R2 = 5 / 361 ≅ 0.014 W.
7. (From S2005 PHY2054 exam) An electron moves at a speed of v = 3 × 107 m/s parallel to and
a distance d = 10 cm from a conducting wire carrying a current of I = 80 A. If the directions of
the electron and current are as shown in the figure, what is the magnitude (in N) and direction of
the force experienced by the electron?
(1) 7.7 × 10−16 toward the wire
(2) 3.4 × 10−16 away from the wire
(3) 1.8 × 10−16 away from the wire
(4) 6.1 × 10−17 toward the wire
(5) 2.6 × 10−16 toward the wire
I
d
e−
The B field a distance d from the wire is B = μ0 I / 2π d , so the force on the electron is thus
F = evB = evμ0 I / 2π d . Plugging in the numbers yields F = 7.7 × 10−16 N, and the right hand
rule (accounting for negative charge) shows the force is towards the wire (as expected since we
effectively have two currents in the same direction).
8. (From S2006 PHY2054 exam) A uniform magnetic field B is directed into the page. Charged
particle Q, moving in the plane of the page in the direction indicated, executes the spiral path as
shown. A reasonable explanation for the particle's
X X X X X X X X
behavior is:
X X X X X X X X
(1) the charge is positive and slowing down
X X X X X X X X
(2) the charge is negative and slowing down
X X X X X X X X
(3) the charge is negative and speeding up
X X X X X X X X
X X X X X X X X
X X X X X X X X
3
X X X X X X X X
Q
PHY2049 Fall 2006
(4) the charge is positive and speeding up
(5) the charge is positive and speed is constant
The right hand rule shows that Q is positive (force bends it to the left of its motion) and shrinking
radius of curvature shows that the particle has to be slowing down.
9. (WebAssign 28.48) A circular wire of radius 10 cm carries a current of 2 A. The loop is placed
so that the normal to its plane makes an angle of 45° with a uniform magnetic field of magnitude
8 T. What is the magnitude of the torque acting on the loop in N⋅m?
(1) 0.36
(2) 0.19
(3) 1.2
(4) 5.0
(5) 7.1
The magnitude of the torque experienced by a current loop is τ = μ B sin θ = π r 2 IB sin θ , where
μ = π r 2 I is the magnetic moment of the loop, B is the magnetic field and θ is the angle between
the plane normal and the B field. This yields τ = 0.36 N⋅m.
10. (WebAssign 29.20) As shown in part (a) of the figure, current is set up in a wire loop consisting of a semicircle of radius 10 cm, a smaller concentric semicircle, and two radial straight
lengths, all in the same plane. The magnitude of the magnetic field produced at the center
marked by the dot is 10 μT. The smaller semicircle is then flipped over as shown in part (b). The
magnetic field produced at the same center marked by the dot now has magnitude 50 μT, and its
direction is reversed. What is the radius of the smaller semicircle in cm?
(1) 6.7
(2) 8.1
(3) 4.3
(4) 2.6
(5) 1.7
(a)
(b)
The B field from a semicircular current (radius R and current I) is B = μ0 I / 4 R . If a is the outer
radius and b is the inner radius, then the B field for (a) is Ba = 14 μ0 I (1/ a − 1/ b ) and the B field
for (b) is Bb = 14 μ0 I (1/ a + 1/ b ) . The ratio of (b) to (a) is Bb / Ba = (1/ a + 1/ b) ) / (1/ a − 1/ b ) or
Bb / Ba = ( a + b ) / ( b − a ) . Using Bb / Ba = −5 , we obtain b = 2a/3 = 6.7 cm.
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