Halliday 8 ed 3 vol cap 28

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1. (a) The force on the electron is

(

)(

)

ˆ
i
j
i
FB = qv × B = q vx ˆ + v y ˆ × Bx ˆ + By j = q ( vx By − v y Bx ) k

(
(

)(
)

)

(

)

= −1.6 ×10−19 C ⎡ 2.0 × 106 m s ( −0.15 T ) − 3.0 × 106 m s ( 0.030 T ) ⎤


ˆ
= 6.2 × 10−14 N k.

Thus, the magnitude of FB is 6.2 × 1014 N, and FB points inthe positive z direction.
(b) This amounts to repeating the above computation with a change in the sign in the
charge. Thus, FB has the same magnitude but points in the negative z direction, namely,
ˆ
F = − 6.2 × 10−14 N k.
B

(

)

2. (a) We use Eq. 28-3:
FB = |q| vB sin φ = (+ 3.2 × 10–19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10–18 N.
(b) a = FB/m = (6.2 × 10– 18 N) / (6.6 × 10–27 kg) = 9.5 × 108 m/s2.
(c) Since it is perpendicular to v , FB does not do any work on the particle. Thus from the
work-energy theorem both the kinetic energy and the speed of the particle remain
unchanged.

3. (a) Eq. 28-3 leads to
v=

6.50 × 10−17 N
FB
=
= 4.00 × 105 m s .
−19
−3
eB sin φ
.
160 × 10 C 2.60 × 10 T sin 23.0°

c

hc

h

(b) The kinetic energy of theproton is
K=

2
121
mv = (1.67 ×10−27 kg ) ( 4.00 × 105 m s ) = 1.34 ×10−16 J ,
2
2

which is equivalent to K = (1.34 × 10– 16 J) / (1.60 × 10– 19 J/eV) = 835 eV.

4. The force associated with the magnetic field must point in the j direction in order to

cancel the force of gravity in the − j direction. By the right-hand rule, B points in the

ej

− k direction (since i × − k = j ).Note that the charge is positive; also note that we need
to assume By = 0. The magnitude |Bz| is given by Eq. 28-3 (with φ = 90°). Therefore, with
m = 1.0 × 10−2 kg , v = 2.0 ×104 m/s and q = 8.0 × 10−5 C , we find
⎛ mg ⎞ ˆ
ˆ
ˆ
B = Bz k = − ⎜
⎟ k = (−0.061 T)k
qv ⎠


5. Using Eq. 28-2 and Eq. 3-30, we obtain

d

i

db g

i

F = q v x By − v y Bx k = q v x 3Bx − v y Bx kwhere we use the fact that By = 3Bx. Since the force (at the instant considered) is Fz k
where Fz = 6.4 × 10–19 N, then we are led to the condition
q ( 3vx − v y ) Bx = Fz ⇒

Bx =

Fz
.
q ( 3vx − v y )

Substituting vx = 2.0 m/s, vy = 4.0 m/s and q = –1.6 × 10–19 C, we obtain
Fz
6.4 × 10−19 N
Bx =
=
= −2.0 T.
q(3vx − v y ) (−1.6 ×10−19 C)[3(2.0 m/s) − 4.0 m]

6. The magnetic forceon the proton is
F = qv × B
where q = +e . Using Eq. 3-30 this becomes
(4 × 10−17 )i + (2 × 10−17)j = e[(0.03vy + 40)i + (20 – 0.03vx)j – (0.02vx + 0.01vy)k]
^

^

^

^

with SI units understood. Equating corresponding components, we find
(a) vx = −3.5×103 m/s, and
(b) vy = 7.0×103 m/s.

^

7. Straight line motion will result from zero net force acting on the system; we ignoregravity. Thus, F = q E + v × B = 0 . Note that v ⊥B so v × B = vB . Thus, obtaining the

d

i

speed from the formula for kinetic energy, we obtain
B=

E
E
100 V /(20 ×10−3 m)
=
=
= 2.67 ×10−4 T.
3
−19
−31
v
2 K / me
2 (1.0 ×10 V ) (1.60 ×10 C ) / ( 9.11×10 kg )

ˆ
In unit-vector notation, B = −(2.67 ×10−4 T)k .

d

i

8. Letting F = q E + v × B = 0 , we get vB sin φ= E . We note that (for given values of
the fields) this gives a minimum value for speed whenever the sin φ factor is at its
maximum value (which is 1, corresponding to φ = 90°). So

vmin =

E 1.50 ×103 V/m
=
= 3.75 ×103 m/s .
B
0.400 T

d

i

9. We apply F = q E + v × B = me a to solve for E :
E=

me a
+ B×v
q

c9.11 × 10
=
e

−31

hd

i + b400µTgi × b12.0 km sg j+ b15.0 km sgk

kg 2.00 × 1012 m s i

.
−160 × 10

2

−19

C

j

.
= −114 i − 6.00 j + 4.80k V m .

10. (a) The net force on the proton is given by
ˆ
F = FE + FB = qE + qv × B = (1.60 × 10−19 C ) ⎡( 4.00 V m ) k+ ( 2000 m s ) ˆ × ( −2.50 × 10−3 T ) ˆ ⎤
j
i


−18
ˆ
= (1.44 × 10 N ) k.
(b) In this case

F = FE + FB = qE + qv × B

(
(

)
)

ˆ
= 1.60 ×...
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