Download 750 CHAPTER 20. MAGNETIC FORCES AND THE MAGNETIC

750
20.16
CHAPTER 20.
MAGNETIC FORCES AND THE MAGNETIC FIELD
The time interval t the particle spends in the region of the field is
t=
distance travelled in the field
.
speed in the field
Since the magnetic force is always perpendicular to the velocity, the particle travels in a circular path at a
constant speed.
Use Newton’s second law to find the radius of the particle’s trajectory in the field. The magnetic force
on a charge in a magnetic field is
~
~
F = q~
v × B,
which has magnitude
F = |q|vB sin θ.
Here θ = 90◦, so
F = |q|vB sin 90◦ = |q|vB.
Now apply Newton’s second law to the particle, using the magnitudes of the vectors. Since the acceleration
is the centripetal acceleration, we have
F = ma =⇒ |q|vB =
mv
mv2
=⇒ r =
.
r
|q|B
Now put this value for r in the above equation for t, noting that the length of the particle’s trajectory within
the region of the field is half the circumference of the circle, πr,
t=
=⇒ t =
πr
distance
=
speed
v
πm
π mv
=
.
v |q|B
|q|B
20.17
a) The electric force on each charge is
~
~
Felec = q E.
Since the particles are negatively charged, the direction of the electric force is opposite to that of the electric
field. This is shown in the sketch below.
s
f
~
Fmagnet 6
−j
~
Felec
?
s
f
~
v
s
f
s
f
sf 6 sf
~
E field
sf
sf
sf
sf
sf
sf
sf
~ field
B
(out of page)
sf
sf
sf
In a velocity selector, the direction of the magnetic force must be opposite to that of the electric force, in
order for the two forces to sum to zero for the selected velocity. Hence the direction of the magnetic force
on the particles must be as shown above.
751
20.18.
b)
The magnetic force on a charge in a magnetic field is
~
~
v × B.
Fmagnet = q~
~ Hence we want
v × B.
Fmagnet is opposite to the direction of ~
In our case, q is negative, so the direction of ~
~
~
~
v × B to be opposite to the direction of Fmagnet . Play with the vector product right-hand rule and you will
~ to point out of the page, as shown in the above sketch.
see that we need B
c) For the selected morons, the electric and magnetic forces vector sum to zero, so their magnitudes must
be equal. Hence, for the selected morons,
|q|E = |q|vB sin θ = |q|vB sin 90◦ = |q|vB =⇒ v =
E
2.00 × 104 N/C
=
= 1.3 × 105 m/s .
B
0.15 T
20.18
a) The magnetic force on a charge in a magnetic field is
~
~
F = q~
v × B.
~ Use the vector product
For a positive charge q, the direction of ~
F is the same as the direction of ~
v × B.
~
right-hand rule to determine the initial direction of ~
v × B. Since the magnetic force always is perpendicular
to the velocity, the particles follow a circular trajectory in the magnetic field, moving with a constant speed
in the clockwise direction as they enter the magnetic field shown in Figure P. 18 on page 939 of the text.
s
f
+j
↑
d
↓
sf
sf
sf
s
f
sf
sf
sf
~
vs
f
sf
sf
sf
~
B field
~
Fmag
v sf (outsfof page)
s ~
f
sf
?
b) Since the conservative electric force is the only force accelerating the particle from rest, the work done
by nonconservative forces is zero. Hence, during the acceleration, the CWE theorem is
Wnonconservative = ∆(KE + PE)
1 2
mv + q(0 V ) − (0 J + qV )
=⇒ 0 J = (KE + PE)f − (KE + PE)i =
2
=⇒ v =
c)
r
2qV
.
m
Within the magnetic field, the magnitude of the magnetic force is (recall that |q| is positive)
F = |q|vB sin θ = |q|vB sin 90◦ = |q|vB.
The particles follow a circular path, experiencing a centripetal acceleration. Apply Newton’s second law,
using the magnitudes of the vectors,
F = ma =⇒ |q|vB =
mv2
.
r