Download Exam 2 Solutions

PHY2054 Spring 2014
Prof. Paul Avery
Prof. Pradeep Kumar
Mar. 18, 2014
Exam 2 Solutions
1. A series circuit consists of an open switch, a 6.0 Ω resistor, an uncharged 4.0 µF capacitor and
a battery with emf 15.0 V and internal resistance 0.5 Ω. Just after the switch is closed, what is
the power dissipation in the battery?
Answer: 2.7 W
Solution: Just after the switch is closed, the capacitor is uncharged so the current is governed
solely by the total resistance in the circuit, yielding a current of i = 15 / 6.5 = 2.31 A. The power
dissipation by the battery’s internal resistance is then P = i 2r = 2.312 × 0.5 = 2.7 W.
2. A high-voltage transmission line delivers 800 MW of power at 400 kV to a consumer area
over a 40 mile total distance. If the resistance in the wire is 0.08 Ω/mile, what is the approximate
power loss due to ohmic heating of the line?
Answer: 13 MW
Solution: The heating is caused by i 2 R heating. The current is obtained easily from the power
equation i = P / V = 800 × 106 / 400 × 103 = 2000 A. With R = 0.08 × 40 = 3.2Ω this gives a
power dissipation of 12.8 MW.
3. A current of 3.0 A flows around the square loop (side length = 40 cm) immersed in a uniform
magnetic field B = 0.5T as shown. The normal to the loop is oriented 90° to the direction of the
B field. The net magnetic force acting on the loop is (in N)
Answer: 0
Solution: In a uniform magnetic field, the forces acting on a current loop always cancel. To have
a net force the loop would have to be in a non-uniform field.
4. A particle of mass 5.0 × 10−21 kg moves in a circular path perpendicular to a uniform magnetic
field of magnitude B = 2.4T. If the orbital frequency is 2.5 × 106 revolutions/s, what is the
magnitude of the particle's charge, in C?
Answer: 3.3 × 10−14
Solution: The centripetal force acting on a charged particle in a circular orbit is provided by the
magnetic force, or mv 2 / R = qvB . This yields R = mv / qB . The frequency f is the inverse of the
period T or 1/ f = T = 2π R / v = 2π m / qB . Solving for q yields q = 3.3× 10−14 C.
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PHY2054 Spring 2014
5. Four long parallel wires are arranged on a plane, as shown in
the figure, with 2.0 cm gaps between them. Each wire carries a
12.0 A current in the direction indicated by the arrow. On the
wire labeled C, what is the magnetic force per meter?
Answer: 3.6 × 10−3 N/m
Solution: From the current directions, we see that all the
magnetic forces point in the upward direction. Each force per unit length is given by
A
B
C
D
F / L = µ0i 2 / 2π d where i is the common current and d is the distance between two wires.
Using µ0i 2 / 2π = 2.88 × 10−5 , we find 2.88 × 10−5 (1/ 0.04 + 1/ 0.02 + 1/ 0.02 ) = 3.6 × 10−3 N/m.
6. The figure shows a mass spectrometer in which charged
particles of different masses enter a region through a slit and
move perpendicular to a uniform magnetic field permeating the
region (the gray dots represent the magnetic field pointing out
of the page). For each particle, the detector measures the
distance between the entrance point to the place where it strikes
the bottom of the region. Assume 12C and 16O singly charged
ions are accelerated to the same velocity before entering the
spectrometer. If the 16O ions strike the detector 5.0 m from the
slit, at one point will the 12C ions hit it?
Answer: 3.75 m
Solution: From the equation equating centripetal force to the magnetic force, mv 2 / R = qvB , we
find R = mv / qB . Thus for the same velocity ions will strike the surface at a distance
proportional to the mass. The distance for 12C ions is then d = (12 / 16 ) 5.0 = 3.75 m.
7. Two long ideal solenoids (with radii 20 mm and 40 mm, respectively) have the same number
of turns of wire per unit length. The smaller solenoid is mounted inside the larger, along a
common axis. The magnetic field within the inner solenoid is zero. If the current in the outer
solenoid is I, what is the current in the inner solenoid?
Answer: I
Solution: The magnetic field of a solenoid is B = µ0 In , where n is the number of windings per
meter and I is the current. Therefore B is independent of radius and the inner solenoid cancels the
B field of the outer solenoid with the same current I.
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PHY2054 Spring 2014
8. Four long straight wires, each with current 12 A, overlap to
form a 12 cm × 24 cm rectangle, as shown. What is the
magnitude of the magnetic field at the point P, midway between 1
the vertical wires and 6 cm above the upper wire?
Answer: 2.7 × 10−5 T
3
4
P
2
Solution: The B fields from the two vertical wires cancel and
the B field from the two horizontal wires partially cancel, with
one distance 3 times the other. The total B field magnitude is therefore B = µ0i / 2π d (1− 1/ 3)
(
)
with d = 0.06 m. Plugging in values gives B = 2.7 × 10−5 T.
9. Three concentric circular current loops of radius r, 2r and 3r carry currents
in the directions shown. The total B field at the center is 0. If the current in
each of the outermost loops is I = 15 A, what is the magnitude of the current
I I
in the innermost loop?
Answer: 12.5 A
Solution: The B field magnitude at the center of a circular loop is
B = µ0i / 2R , where R is the radius and i is the current. From the directions of the currents and
the relation of the loop radius, the B = 0 condition gives 0 = I / 3+ I / 2 − I1 , where Il is the
current in the innermost loop. This yields I1 = 5I / 6 = 12.5 A.
10. A long wire carries a 500 A current as shown. A particle with
charge −40 mC moves towards the wire along a perpendicular
path at a velocity of 2.0 × 104 m/s. What is the magnitude of the
force (in N) on the particle when it is 12 cm from the wire?
Answer: 6.7 × 10−1
Solution: The magnitude of the force is F = qvB with B = µ0i / 2π d (since the particle is
moving perpendicular to the B field). Substituting yields the total force of 0.67 N.
11. A series LR circuit consists of a battery, resistor, inductor and a switch. When the switch is
closed the circuit has a time constant of 8.0 s. If the resistance and inductance of the circuit are
both doubled, what is the resulting time constant?
Answer: 8.0 s
Solution: The time constant is L / R. If both are doubled the time constant is unchanged.
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PHY2054 Spring 2014
12. An electron with velocity 8.0 × 105 m/s moves 30° relative to the direction of a 1.5 T
magnetic field. What is the force on the electron in N?
Answer: 9.6 × 10−14
Solution: The magnitude of the force is given by F = evBsin φ where φ = 30° . Plugging in
values gives F = 9.6 × 10−14 N.
13. A series LR circuit includes a 9.0 V battery, a resistance of 0.5 Ω, and an inductance of 0.8 H.
What is the greatest energy stored by the inductor after the switch is closed?
Answer: 130 J
Solution: The current will increase to its maximum value governed only by the resistance, or
i = V / R = 18 A. The energy stored in the solenoid is then U L = 12 Li 2 = 0.5 × 0.8 × 182 = 128.8 J.
14. In the RL circuit shown, L = 7.0 mH, R1 = 10.0 Ω and
R2 = 5.0 Ω and V = 18.0 V (DC). What is the voltage
across R2 immediately after the switch is closed?
Answer: 6.0 V
Solution: Just after the switch is closed, there is no
current in the inductor so the current is governed solely by
the resistors. The voltage across R2 is then 5 / 15 = 1 / 3 of
the total or 6.0 V.
V
+
R1
L
R2
S
15. Refer to the preceding problem. After the switch has been closed for a very long time, it is
opened. What is the voltage across R2 immediately after the switch is opened?
Answer: 9.0 V
Solution: After the switch has been closed for a long time, the current in the inductor rises to its
maximum value of V / R1 = 18 / 10 = 1.8 A (no current moves across R2). Just after the switch is
opened, the current remains at that value (decreasing exponentially afterwards), thus the voltage
across R2 is 1.8 × 5.0 = 9.0 V.
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PHY2054 Spring 2014
16. A 1000-turn circular coil with a radius of 20 cm is mounted on a rotating frame that rotates at
10 turns/sec in the presence of a 0.10-T uniform magnetic field that is perpendicular to the axis
of rotation. The total resistance of the coil is 10 Ω. What is the approximate rms current in the
coil?
Answer: 56 A
Solution: The instantaneous emf in a rotating loop from Faraday’s law is N ω Asin ω t , where N
is the number of windings, A is the area of the loop and ω = 2π f is the angular frequency. The
rms value of the current is therefore ( N ω A / R ) / 2 , which yields 56 A as the rms current.
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17. As shown in the figure, a conducting rod of length 40 cm, resistance
0.20 Ω and mass 100 g slides along frictionless conducting rails in a
constant uniform magnetic field B = 0.60 T perpendicular to the rails. The
rails have negligible electrical resistance. The rod is pulled down by gravity
(g = 9.8 m/s2). After a long time, what is the speed of the rod?
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Answer: 3.4 m/s
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Solution: Current flows in a loop formed by the rails and the bar on account
of the induced emf, or i = BLv / R . The magnetic force is then F = ILB . After a long time the
magnetic force is the same as the gravitational force or mg = ILB = ( BLv / R ) LB . Solving for the
velocity yields 3.4 m/s.
18. A solenoid of length 75 cm and diameter 25.0 cm is wound with 20 turns per cm of wire. In
the middle of the solenoid is a 40-turn wire loop of radius 6.0 cm oriented with its normal axis
lying along the solenoid axis. If the solenoid current is increasing at a rate of 100 A/s, what is the
induced emf in the wire loop (in V)?
Answer: 0.11
Solution: The solenoidal B field is B = µ0in , where i is the current and n is the number of wire
turns per meter. From Faraday’s law, the induced current in the loop is
ε = N Δφ B / Δt = N π r 2 ( µ0 nΔi / Δt ) , where N is the number of turns in the wire loop, r is the
radius of the loop and Δi / Δt is the rate of change of the current in the solenoid. Multiplying the
terms yields an induced emf of 0.11 V.
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PHY2054 Spring 2014
19. An electrical device with an equivalent resistance of 100 Ω has a maximum instantaneous
voltage rating of 60 V. Assume that it is plugged into a standard 120 V electrical outlet in series
with a resistance R. What is the approximate minimum value of R needed to keep the
instantaneous voltage across the device within its safety limit?
Answer: 180 Ω
Solution: Let ε max = 120 2 = 169.7 V be the maximum emf of the electrical outlet. The
maximum voltage across the device is then Vmax = 60 = ε max 100 / (100 + R ) . Solving for R yields
182 Ω.
20. Two 100 W bulbs differ in that one is designed for the American 120 V rms (60 Hz) and the
other is designed for the European 220 V rms (50 Hz). Which bulb has the greater resistance, and
what is that resistance in ohms?
Answer: European, 484
2
/ R , independent of frequency. Solving for R gives 484 Ω.
Solution: The power is P = Vrms
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