Download Exam 2 Solutions

PHY2054 Spring 2008
Prof. P. Kumar
Prof. P. Avery
March 5, 2008
Exam 2 Solutions
1. Two cylindrical resistors are made of the same material and have the same resistance. The
resistors, R1 and R2, have different radii, r1 and r2, and different lengths, L1 and L2. Which of the
following relative values for radii and lengths would result in equal resistances?
(1) 2r1 = r2 and 4L1 = L2
(2) 2r1 = r2 and L1 = 2L2
(3) r1 = r2 and 4L1 = L2
(4) r1 = r2 and L1 = 2L2
(5) none of these
The resistance is given by R = ρ L / A , where ρ is the resistivity, L is the length and A is the cross
sectional area. To keep R unchanged, the ratio of L to A must be unchanged. Since A = πr2, if r
is doubled then L must be multiplied by 4. Only the answer (1) works.
2. If a 9.0-V battery with internal resistance 0.2 Ω and an 18 Ω resistor are connected together,
what is the amount of electrical energy (in J) transformed to heat per coulomb of charge that
flows through the circuit?
(1) 9.0
(2) 3.0
(3) 0.50
(4) 72
(5) 4.5
The amount of heat released is PΔt, where P = Vi. The total amount of charge is q = iΔt. Thus
the amount of heat energy released per coulomb of charge is PΔt / iΔt = P/i = V = 9V.
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PHY2054 Spring 2008
3. (2054 exam, #1, 2006) At room temperature conditions of 25° C, a circuit containing a metal
resistor connected to a 50 V battery has a current of 2.00 A. When the wire is heated to 145° C
the current drops to 1.94 A. What is the temperature coefficient of resistivity of this metal?
(1) 0.00026 / °C
(2) 0.0064 / °C
(3) 0.031 / °C
(4) 0.00075 / °C
(5) 0.077 / °C
The resistance at 25° C is 50 / 2.00 = 25 Ω. The resistance at 145° C is 50 / 1.94 = 25.773 Ω.
Thus the temperature coefficient of resistivity is 0.773 / 25.0 / 120° = 0.00026 / °C.
4. (Chap. 18.9) Consider the circuit shown here, with R = 15 Ω. What is the potential difference
between points a and b?
10
25 V
(1) 5.56 V
(2) 19.4 V
(3) 25 V
(4) 12.3 V
(5) None of these
a
b
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5
This circuit is equivalent to a 10 Ω resistor in series with a parallel combination of 10 Ω, 5 Ω
and 20 Ω, yield a total resistance of 12.86 Ω. Thus the total current is 25 / 12.86 = 1.944 A.
The voltage difference is thus 25 – 1.944*10 = 5.56 V.
5. (2049 Exam 2, #4, Fall ’07) Consider the circuit shown in the accompanying figure. What is
the power dissipated (in watts) in the 5 Ω resistor?
(1) 0.43
(2) 0.11
(3) 0.28
(4) 2.35
(5) 0.069
i1
i2
The two equations from Kirchoff’s rules are 1 − 2i1 − 5 ( i1 + i2 ) = 0 and 2 − i2 − 5 ( i1 + i2 ) = 0 .
Solving yields i1 + i2 = 5 /17 = 0.294 A. So the power is 0.2942 × 5 = 0.43 W.
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PHY2054 Spring 2008
6. (Test Bank 6) When a 24.0-Ω resistor is connected across a 12.0-V battery, a current of 482
mA flows. What is the internal resistance of the battery?
(1) 0.9 Ω
(2) 0.02 Ω
(3) 25.0 Ω
(4) 49.8 Ω
(5) 0.008Ω
The total resistance is R = 12.0 / 0.482 = 24.90 Ω. Thus the internal resistance is 0.9 Ω.
7. (Test Bank 27) Two resistors of values 6.0 and 12.0 Ω are connected in parallel. This
combination in turn is hooked in series with a 2.0 Ω resistor and a 24 V battery. What is the
current in amps in the 6 Ω resistor?
(1) 2.7
(2) 4.0
(3) 6.0
(4) 1.3
(5) 8.0
The parallel combination yields a resistance of 4 Ω and thus the total resistance is 6.0 Ω. The
total current is thus 24 / 6.0 = 4 A. To find the current in the 6 Ω branch, first find the voltage
across it, which is the same as the voltage across the parallel combination. This voltage is
determined from 24 – 4*2 = 16 V. Thus the current is 16 / 6 = 2.67 A.
8. (Test Bank 45) What is the current through the 8 Ω resistor?
(1) 1.5 A
(2) 0.50 A
(3) 1.0 A
(4) 2.0 A
(5) none of these
Let i1 be the current flowing to the right in the middle branch, and i2 be the current flowing to the
left in the lower branch. The two equations from Kirchhoff’s rule are 16 – 4(i2 – i1) +12i1 – 18 =
0 and 18 – 12i1 – 8i2 = 0. We are solving for i2 only, which yields i2 = 1.5 A. 3
PHY2054 Spring 2008
9. (2054 exam 2 #6, Spring 2007) In the circuit shown, the internal resistance of the battery is
0.1 Ω and the capacitors are initially uncharged. The current (in A) flowing in the 6 Ω resistor an
instant after the switch is closed
is
(1) 2.5
(2) 4.9
(3) 3.9
(4) 3.3
(5) zero
Just after the switch is thrown, the capacitors have no effect. The current is therefore calculated
from 20 / 8.1 = 2.5 A.
10. In the circuit shown, the switch S is open and the capacitor (C = 2.5 μF) is initially
uncharged. The switch is then closed long enough to allow the capacitor to fully charge. If the
switch is now opened, how much time (in μsec) is required for the potential difference across the
capacitor to reach 2 V?
(1) 27
(2) 1.9
(3) 5.2
(4) 40
(5) 22
When the capacitor is fully charged, the voltage across it is 15 V. After the switch is opened, it
discharges through a resistor combination equivalent to 5.33 Ω. The equation describing the
voltage on the capacitor is 15exp ( −t / RC ) = 2 , which yields t = ( 5.33)( 2.5 ) ln (15 / 2 ) = 27 μs .
11. (Test Bank 19) If a proton is released at the equator and falls toward the Earth under the
influence of gravity, in which direction is the magnetic force on the proton?
(1) east
(2) south
(3) north
(4) west
(5) up
The field lines point north to the geomagnetic south pole at the north geographic pole. Using the
right hand rule yields a direction of east.
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PHY2054 Spring 2008
12. (Test Bank 30) A horizontal current-carrying wire of length 50 cm and mass 0.2 kg is placed
perpendicular to a uniform horizontal magnetic field of magnitude 0.27 T. For what current will
the wire float due to the magnetic force acting upward on it? (g = 9.8 m/s2)
(1) 14.5 A
(2) 1.5 A
(3) 3.9 A
(4) 25 mA
(5) 150 mA
Balance of gravitational and magnetic forces leads to the equation mg = BiL. Solving for i
yields 14.5 A.
13. (Test Bank 39) A rectangular coil (0.20 m × 0.80 m) has 200 turns and is placed in a uniform
magnetic field of 0.30 T. If the orientation of the coil is varied through all possible positions, the
maximum torque on the coil by magnetic forces is 0.080 N⋅m. What is the current in the coil?
(1) 8.3 mA
(2) 1.7 A
(3) 2.5 mA
(4) 1.0 A
(5) 50 mA
The maximum torque on the coil is τ = NAiB , where N = 200, A = 0.3 × 0.4 = 0.12 m2, τ =
0.080 and B = 0.30 T. Solving for i yields 8.3 mA.
14. (Chap. 19.31) A singly charged positive ion has a mass of 2.68 × 10-26 kg. After being
accelerated through a potential difference of 232 V, the ion enters a magnetic field of 0.590 T, in
a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.
(1) 1.49 cm
(2) 1.77 cm
(3) 8.82 mm
(4) 4.5 cm
(5) none of these
The radius is found by equating the centripetal force to the magnetic force, or mv 2 / r = evB ,
which yields r = mv / eB . From the acceleration over a potential, we can also write 12 mv 2 = eV .
Putting these together yields the equation r = 2mV / e / B . Solving for r yields 1.49 cm.
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PHY2054 Spring 2008
15. (Chap. 18.35) A lightning bolt may carry a current of 1.00 × 104 A for a short time. What is
the resulting magnetic field 190 m from the bolt in μT? Suppose that the bolt extends far above
and below the point of observation.
(1) 10.5
(2) 20.00
(3) 120
(4) 66.1
(5) none of these
The B field at a distance r from a long straight wire, which is what the lightning bolt essentially
(
)( )
is, is given by B = μ0i / 2π r = 2 × 10−7 104 /190 = 1.05 × 10−5 T = 10.5 μT.
16. (PHY 2049 exam 2 #9, Fall 2007) Four long parallel wires are arranged on a plane, as
shown in the attached figure, with 2.0 cm gaps between them. Each wire carries a 6.0 A current
in the direction indicated by the arrow. On the wire labeled B, what is the magnetic force per
meter in N/m?
(1) 1.8 × 10−4
(2) 2.8 × 10−4
(3) 6.0 × 10−3
(4) 4.5 × 10−3
(5) 9.0 × 10−5
We first note that wire B is repelled from both A and C, so there is no net effect from them. The
only net force is with D. A quick calculation yields the value of the force per unit length as
(
)
F / L = μ0i AiC / 2π d = 4π × 10−7 × 6 × 6 / ( 2π × 0.04 ) = 1.8 × 10−4 N/m.
17. (PHY2054 exam 2, #12, Spring 2006) A long straight wire and a rectangular conducting loop
lie in the same plane as shown. If the current in the long wire is decreasing, what is the direction
of the net force exerted on the loop?
(1) up
(2) down
(3) out of the page
(4) into the page
(5) zero
The induced current in the loop is clockwise from Lenz’ law. From the right hand rule, the force
on the current in the upper branch is up and on the lower branch is down. However, the B field
is stronger at the upper branch, so the net force is up. (The forces on the left and right branches
cancel because each element of each branch encounters the exact same B field.)
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PHY2054 Spring 2008
18. (Chap. 20.18) Consider the arrangement shown in the figure. Assume that R = 4.0 Ω and L =
1.2 m, and that a uniform 2.8 T magnetic field is directed into the page. The bar is moving at a
speed of 0.714 m/s. What is the current in amps through the resistor R?
(1) 0.6
(2) 0.5
(3) 0.4
(4) 0.7
(5) 0.3
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The emf is given by E = BLv, so the current is BLv / R, yielding 0.6 A.
19. (Chap. 20.12) A 480 turn circular-loop coil 7.25 cm in radius is initially aligned so that its
axis is parallel to the Earth's magnetic field. In 2.77 ms, the coil is flipped so that its axis is
perpendicular to the Earth's magnetic field. If an average voltage of 0.170 V is thereby induced
in the coil, what is the value of the Earth's magnetic field (in μT) at that location?
(1) 59
(2) 48
(3) 65
(4) 39
(5) 43
The emf for a circular coil which is rotated in this way is E = N π r 2 B / Δt , where N = 480, r =
0.0725, E = 0.170 and Δt = 0.00277. This yields B = 59 μT.
20. (Test Bank 14) A rectangular loop of wire (3 cm × 4 cm) is placed inside a solenoid of
radius 8 cm. The solenoid is 25 cm long and is wound with 1500 turns of wire carrying 4 A of
current. The normal to the loop lies along the solenoid axis. If the current in the solenoid is
reduced to zero in 0.2 sec., find the magnitude of the average induced emf in the loop.
(1) 0.180 mV
(2) 36 μV
(3) 1.8 μV
(4) 0.25 mV
(5) 80 μV
The B field strength of a solenoid is B = μ0in , where n = 1500 / 0.25 = 6000. The emf in the
loop can then be determined from E = Aμ0 nΔi / Δt , where A = 0.03 × 0.04 = 0.0012, Δi = 4 and
Δt = 0.2. Solving for the emf yields E = 0.18 mV.
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