1137
Chapter 28 (# 6, 7, 11, 16, 26, 33, 79, 84)
6. An electric field of 1.50 kV/m and a magnetic field of 0.400 T act on a moving electron to produce no net force. If the fields are perpendicular to each other, what is the electron’s speed?
6. Letting , we get vB sin f = E. We note that (for given values of the fields) this gives a minimum value for speed whenever the sin f factor is at its maximum value (which is 1, corresponding to f = 90°). So
vmin = E / B = (1.50 ´ 103 V / m) / (0.400 T) = 3.75 ´ 103 m / s.
7. In Fig. 28-33, an electron accelerated from rest through potential difference V1 = 1.00 kV enters the gap between two parallel plates having separation d = 20.0 mm and potential. Neglect fringing and assume that the electron’s velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap?
7. Straight line motion will result from zero net force acting on the system; we ignore gravity. Thus, . Note that so . Thus, obtaining the speed from the formula for kinetic energy, we obtain
In unit-vector notation, .
11. A metal strip 6.50 cm long, 0.850 cm wide, and 0.760 mm thick moves with constant velocity v through a uniform magnetic field B = 1.20 mT directed perpendicular to the strip, as shown in Fig. 28-35. A potential difference of 3.90 µV is measured between points x and y across the strip. Calculate the speed v.
11. For a free charge q inside the metal strip with velocity we have . We set this force equal to zero and use the relation between (uniform) electric field and potential difference. Thus,
16. An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
16. (a) The accelerating process may be seen as a conversion of potential energy eV into kinetic energy. Since it starts from rest, and
(b) Eq. 28-16 gives
26. In Fig. 28-38, an electron with an initial kinetic energy of 4.0 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.010 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 25.0 cm. There is an electric potential difference ΔV = 2000 V across the gap, with a polarity such that the electron’s speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude 0.020 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?
26. Eq. 28-17 gives T = 2pme /eB. Thus, the total time is
+ tgap + = + tgap .
The time spent in the gap (which is where the electron is accelerating in accordance with Eq. 2-15) requires a few steps to figure out: letting t = tgap then we want to solve
d = vo t + a t2
0.25 m = t + t2
for t. We find in this way that the time spent in the gap is t » 6 ns. Thus, the total time is 8.7 ns.
33. A horizontal power line carries a current of 5000 A from south to north. Earth’s magnetic field (60.0 µT) is directed toward the north and inclined downward at 70.0º to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on 100 m of the line due to Earth’s field.
33. (a) The magnitude of the magnetic force on the wire is given by FB = iLB sin f, where i is the current in the wire, L is the length of the wire, B is the magnitude of the magnetic field, and f is the angle between the current and the field. In this case f = 70°. Thus,
(b) We apply the right-hand rule to the vector product to show that the force is to the west.
79. A proton, a deuteron (q = +e, m = 2.0 u), and an alpha particle (q = +2e, m = 4.0 u) are accelerated through the same potential difference and then enter the same region of uniform magnetic field B, moving perpendicular to B. What is the ratio of (a) the proton’s kinetic energy Kp to the alpha particle’s kinetic energy Kα and (b) the deuteron’s kinetic energy Kd to Kα? If the radius of the proton’s circular path is 10 cm, what is the radius of (c) the deuteron’s path and (d) the alpha particle’s path?
79. (a) Since K = qV we have or
(b) Similarly,
(c) Since , we have
(d) Similarly, for the alpha particle, we have
84. A proton, a deuteron (q = +e, m = 2.0 u), and an alpha particle (q = +2e, m = 4.0 u) all having the same kinetic energy enter a region of uniform magnetic field B, moving perpendicular to B. What is the ratio of (a) the radius rd of the deuteron path to the radius rp of the proton path and (b) the radius rα of the alpha particle path to rp?
84. Referring to the solution of problem 19 part (b), we see that implies the proportionality:. Thus,
(a) and
(b)