27: 9, 14, 20, 26, 28 (reminder: all solutions must start with OSE’s)
27.9.IDENTIFY: Apply F  qv  B to the force on the proton and to the force on the electron.
Solve for the components of B and use them to find its magnitude and direction.
SET UP: F is perpendicular to both v and B. Since the force on the proton is in the
 y-direction, B y  0 and B  Bx iˆ  Bz kˆ. For the proton, vp  (150 km/s)iˆ  vpiˆ and
Fp  (225 1016 N) ˆj  Fp ˆj. For the electron, ve  (475 km/s)kˆ  vekˆ and
Fe  (850 1016 N) ˆj  Fe ˆj. The magnetic force is F  qv  B.
EXECUTE: (a) For the proton, Fp  qvp  B gives Fp ˆj  evpiˆ  ( Bxiˆ  Bz kˆ )  evp Bz ˆj. Solving
for
Bz
gives
Bz  
Fp
evp

225 1016 N
(160 1019 C)(1500 m/s)
 09375 T.
For the electron,
which gives Fe ˆj  (e)(vekˆ)  (Bxiˆ  Bz kˆ)  eveBx ˆj. Solving for
Bx 
16
Fe
850  10
N

 1118 T.
eve (160  1019 C)(4750 m/s)
Therefore
Bx
Fe  eve  B,
gives
B  1118 Tiˆ  09375 Tkˆ.
The magnitude
of the field is B  Bx2  Bz2  (1118 T)2  (09375 T)2  146 T. Calling  the angle that the
magnetic field makes with the
 x -axis,
we have
tan  
Bz 09375 T

 08386,
Bx
1118 T
Therefore the magnetic field is in the xz-plane directed at 40.0° from the
the – z -axis, having a magnitude of 1.46 T.
(b) B  Bxiˆ  Bz kˆ and v  (32 km/s)( ˆj).
so   40.0.
 x-axis
toward
F  qv  B  (e)(32 km/s)( ˆj)  ( Bxiˆ  Bz kˆ)  e(32 103 m/s)[ Bx (kˆ)  Bz iˆ].
F  e(32 103 m/s)(1118 Tkˆ  09375 Tiˆ)  480 1016 Niˆ  5724 1016 Nkˆ.
F  Fx2  Fz2  747 1016 N.
have
tan  
Calling  the angle that the force makes with the
Fz 5724  1016 N

,
Fx 4800  1016 N
directed at 50.0° from the
– x -axis,
we
which gives   500. The force is in the xz-plane and is
– x-axis
toward either the
– z-axis.
EVALUATE: The force on the electrons in parts (a) and (b) are comparable in magnitude
because the electron speeds are comparable in both cases.
Physics 2135
Homework Assignment #15 (Tuesday, October 11, 2016)
Page 1
27.14. IDENTIFY: When B is uniform across the surface,  B  B  A  BA cos.
SET UP: A is normal to the surface and is directed outward from the enclosed volume.
For surface abcd, A   Aiˆ. For surface befc, A   Akˆ. For surface aefd, cos   3/5 and the
flux is positive.
EXECUTE: (a)  B (abcd )  B  A  0.
(b)  B (befc)  B  A  (0128 T)(0300 m)(0300 m)  00115 Wb.
r r
(c)  B (aefd )  B  A  BA cos  53 (0128 T)(0500 m)(0300 m)  00115 Wb.
(d) The net flux through the rest of the surfaces is zero since they are parallel to the xaxis. The total flux is the sum of all parts above, which is zero.
EVALUATE: The total flux through any closed surface, that encloses a volume, is zero.
The diagram below may help you visualize surface aefd:
Physics 2135
Homework Assignment #15 (Tuesday, October 11, 2016)
Page 2
27.20. IDENTIFY: The magnetic field acts perpendicular to the velocity, causing the ion to
move in a circular path but not changing its speed.
SET UP:
R
mv
|q|B
and
K  12 mv 2 .
EXECUTE: (a) Solving
K = 5.0 MeV = 8.0 1013 J.
K  1 mv 2
2
for v gives
v  2K/m .
v = [2(8.0 1013 J)/(1.67 10 27 kg)]1/2  3.095 10 7 m/s, which rounds to
(b) Using
R
mv
|q|B
3.1107 m/s.
= (1.67 1027 kg)(3.095 107 m/s)/[(1.602 10 19 C)(1.9 T) ]  0.17 m  17 cm.
EVALUATE: If the hydride ions were accelerated to 20 MeV, which is 4 times the value used
here, their speed would be twice as great, so the radius of their path would also be twice as great.
27.26. IDENTIFY: After being accelerated through a potential difference V the ion has kinetic
energy qV. The acceleration in the circular path is v 2 /R
SET UP: The ion has charge q  e.
EXECUTE: K  qV  eV .
1 mv 2
2
FB  q v B sin  .   90. F  ma
R
 eV
gives
and
v
q vB  m
2eV
2(160  1019 C)(220 V)

 779  104 m/s.
m
116  1026 kg
v2
.
R
mv (116  1026 kg)(779  10 4 m/s)

 6.46  103 m  6.46 mm.
qB
(160  1019 C)(0874 T)
EVALUATE: The larger the accelerating voltage, the larger the speed of the particle and the
larger the radius of its path in the magnetic field.
Physics 2135
Homework Assignment #15 (Tuesday, October 11, 2016)
Page 3
27.28. IDENTIFY: For no deflection the magnetic and electric forces must be equal in
magnitude and opposite in direction.
SET UP: v  E/B for no deflection. With only the magnetic force, q v B  mv2/R.
EXECUTE: (a) v  E/B  (156 104 V/m)/(462 103 T)  338 106 m/s.
(b) The directions of the three vectors v , E , and B are sketched in Figure 27.28.
mv (911  1031 kg)(338  106 m/s)

 417  103 m.
qB
(160  1019 C)(462  103 T)
(c)
R
T
2 m 2 R 2 (417  103 m)


 774  109 s.
qB
v
(338  106 m/s)
EVALUATE: For the field directions shown in Figure 27.28, the electric force is toward
the top of the page and the magnetic force is toward the bottom of the page.
Figure 27.28
Physics 2135
Homework Assignment #15 (Tuesday, October 11, 2016)
Page 4
Physics 2135
Homework Assignment #15 (Tuesday, October 11, 2016)
Page 5