CHAPTER 29
THE MAGNETIC FIELD
Problem
12. A velocity selector uses a 60-mT magnetic field and a 24 kN/C electric field. At what speed will
charged particles pass through the selector undeflected?
Solution
The condition for zero deflection is v  E/B  (24 kN/C)/(0.06 T)  400 km/s.
Problem
20. The Van Allen belts are regions in space where high-energy charged particles are trapped in Earth’s
magnetic field. If the field strength at the Van Allen belts is 010
. G, what are the period and radius of
the spiral path described (a) by a proton with a 1.0-MeV kinetic energy? (b) by a 10-MeV proton?
Solution
The cyclotron frequency, f  qB/2 m, is independent of the particle’s energy, so the orbital period for
protons of either energy is T  1/ f  2 (1.67  10
kg/1.6  10 C)  (10
The radius is given by the result of
Problem 22 (in atomic units, explained in the solution of Problem 19). (a) For a
27
1 MeV proton, r  2 Km /300qB 
19
5
T)  6.56 ms.
2(1)(938)/300(1)(105)  14.4 km. (b) Since r »
K,
the orbital radius for a proton with ten times the kinetic energy is 10 (14.4 km)  45.7 km.
Problem
30. An electron is moving in a uniform magnetic field of 0.25 T; its velocity components parallel and
.  10 6 m/s. (a) What is the radius of the electron’s spiral
perpendicular to the field are both equal to 31
path? (b) How far does it move along the field direction in the time it takes to complete a full orbit
about the field direction?
Solution
(a) The radius depends only on the perpendicular velocity component,
r  mv  /eB  (9.11  1031 kg)(3.1  106 m/s)  (1.6  10 19 C)(0.25 T)  70.6 m. (b) The
distance moved parallel to the field is d  v||T , where T is the cyclotron period (Equation 29-4). Since
v||  v in this case, d  v || (2 m/eB )  2 (70.6  m)  444
 m.
Problem
36. A wire of negligible resistance is bent into a rectangle as shown in Fig. 29-40, and a battery and
resistor are connected as shown. The right-hand side of the circuit extends into a region containing a
uniform magnetic field of 38 mT pointing into the page. Find the magnitude and direction of the net
force on the circuit.
FIGURE 29-40
Problem 36 Solution.
Solution
The forces on the upper and lower horizontal parts of the circuit are equal in magnitude, but opposite in
direction and thus cancel (see Fig. 29-40), leaving the force on the righthand wire,
IB  (E=R)B  (12 V=3 )(0.1 m)(38 mT)  15.2 mN toward the right, as the net force on the circuit.
Problem
52. A satellite with rotational inertia 20 kg  m 2 is in orbit at a height where Earth’s magnetic field
strength is 0.18 G. It has a magnetic torquing system, as described in Example 29-7, that uses a 1000turn coil 30 cm in diameter. What should be the current in the coil if the magnetic torque is to give the
satellite a maximum angular acceleration of 0.0015 s 2 ?
Solution
If we require that the maximum torque from the coil,  max   B  NIAB (from Equations 27-10 and 11),
should yield the desired angular acceleration,
   max /I rot , then the current necessary is
I   max /NAB   I rot /NAB 
(1.5  10 3 s 2 )(20 kg  m 2 )  (10 3 )
1
4
 (0.3 m) 2 (1.8  10 5 T)  23.6 A.
Problem
64. The coil in the loudspeaker of Fig. 29-31 consists of 100 turns of wire, 3.5 cm in diameter. The
magnetic field strength at the coil is 0.64 T. Find the magnitude of the force on the speaker coil when
the current in the coil is 2.1 A.
Solution
In Fig. 29-31(b), the magnetic force on each element of current in the wire has the same magnitude, IdB,
perpendicular
to the page, so the net force on N turns of diameter d is F  coil IBd  NIB d. (The total length of the
wire is assumed
to be the number of identical turns times the circumference of one turn.) For the values given,
F  (100)(21
. A)  (0.64 T) (3.5 cm)  14.8 N.
z