Problem#1
Terminal Speed. A bar of length L = 0.36 m is free to slide without friction on horizontal rails, as shown in Fig. 1. There is a uniform magnetic field B = 1.5 T directed into the plane of the figure. At one end of the rails there is a battery with emf ε = 12 V and a switch. The bar has mass 0.90 kg and resistance 5.0 Ω and all other resistance in the circuit can be ignored. The switch is closed at time t = 0. (a) Sketch the speed of the bar as a function of time. (b) Just after the switch is closed, what is the acceleration of the bar? (c) What is the acceleration of the bar when its speed is 2.0 m/s? (d) What is the terminal speed of the bar?
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| Fig.1 |
Answer:
Apply Newton’s second law to the bar. The bar will experience a magnetic force due to the induced current in the loop. Use a = dv/dt to solve for v. At the terminal speed, a = 0. The induced emf in the loop has a magnitude BLv. The induced emf is counterclockwise, so it opposes the voltage of the battery, ε.
(a) The net current in the loop is
I = (ε - BLv)/R
The acceleration of the bar is
a = F/m = ILB sin 900/m
a = (ε - BLv)LB/mR
dv/dt = (ε - BLv)LB/mR or
∫0v(ε - BLv)-1dv = ∫0t(LB/mR)dt
v = (ε/BL)(1 – eα)
with α = –B2L2t/mR = –(1.5 T)2(0.36 m)2t/(0.90 kg x 5.0 Ω) = –t/15 s
v = [12 V/(1.5 T x 0.36 m)](1 – e–t/15 s)
v = (22 m/s)(1 – e–t/15 s)
The graph of v versus t is sketched in Figure 2. Note that the graph of this function is similar in appearance to that of a charging capacitor.
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| Fig.2 |
(b) Just after the switch is closed, v = 0 and I = ε /R = 2.4 A, F = ILB =1.296 N, and a = F/m = 1.4 m/s2.
(c) When v = 2 m/s,
a = [12 V – (1.5 T)2 x (0.36 m)2 x 2 m/s]/(0.90 kg x 5.0 Ω)
a = 1.3 m/s2
(d) Note that as the speed increases, the acceleration decreases. The speed will asymptotically approach the terminal speed
vasimp = ε/BL = 12 V/(1.5 T x 0.36 m) = 22 m/s, which makes the acceleration zero.
Problem#2
Antenna emf. A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 2.0-m-long rod. The antenna is oriented perpendicular to the earth’s surface. At the equator, the earth’s magnetic field is essentially horizontal and has a value of 8.0 x 10-5 T ignore any changes in B with altitude. Assuming the orbit is circular, determine the induced emf between the tips of the antenna.
Answer:
The gravitational force on the satellite is
F = GmmE/r2
where m is the mass of the satellite and r is the radius of its orbit.
GmmE/r2 = mv2/r
v2 = GmE/r
v2 = (6.67 x 10-11 Nm2/kg2)(5.972 x 1024 kg)/(4 x 105 m + 6.371 x 106 m)
v = 7.665 km/s
Using this v in ε = vBL gives
ε = (8.0 × 10−5 T)(7.665×103 m/s)(2.0 m) = 1.2 V
Problem#3
emf in a Bullet. At the equator, the earth’s magnetic field is approximately horizontal, is directed toward the north, and has a value of 8.0 x 10-5 T. (a) Estimate the emf induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot toward the east. Assume the bullet has a length of 1 cm and a diameter of 0.4 cm and is traveling at 300 m/s. Which is at higher potential: the top or bottom of the bullet? (b) What is the emf if the bullet travels south? (c) What is the emf induced between the front and back of the bullet for any horizontal velocity?
Answer:
The magnetic force pushed positive charge toward the high potential end of the bullet.
(a) ε = BLv = (8 × 10−5 T)(0.004 m)(300 m/s) = 96 μV
Since a positive charge moving to the east would be deflected upward, the top of the bullet will be at a higher potential.
(b) For a bullet that travels south, v and B are along the same line, there is no magnetic force and the induced emf is zero.
(c) If v is horizontal, the magnetic force on positive charges in the bullet is either upward or downward, perpendicular to the line between the front and back of the bullet. There is no emf induced between the front and back of the bullet.
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