Download PWE 19-1: Magnetic Forces on a Proton and an Electron

Example 19-1 Magnetic Forces on a Proton and an Electron
At a location near our planet’s equator, the direction of Earth’s magnetic field is horizontal (that is, parallel to the ground)
and due north, and the magnitude of the field is 2.5 * 1025 T. Find the direction and magnitude of the magnetic force on
a particle moving at 1.0 * 104 m>s if the particle is (a) a proton moving horizontally and due east, (b) an electron moving
horizontally and due east, and (c) a proton moving horizontally in a direction 25° east of north. Recall that a proton has
charge e = 1.60 * 10219 C and an electron has charge 2e.
Set Up
In each case we’ll use Equation 19-1 to find the
magnitude of the force on the moving proton
or electron. The right-hand rule (Figure 19-5)
will tell us the direction of the magnetic force.
Solve
(a) The velocity vector s
v of the proton is
s,
perpendicular to the magnetic field vector B
so in Equation 19-1 the angle u = 90° and
s both lie in a
v and B
sin u = 1. The vectors s
horizontal plane; the magnetic force vector Fs
must be perpendicular to this plane, so Fs is
vertical. The right-hand rule tells us that the
force points vertically upward.
Magnetic force on a moving
charged particle:
F =  q  vB sin u
B
(19-1)
North
Charge on the proton:
q = e = 1.60 * 10
219
F
C
The proton velocity vector is
perpendicular to the magnetic
field vector (u = 90°), so the
magnitude of the magnetic
force on the proton is
B
q = +e
up
v
North
F =  q  vB sin u = evB sin u
= 11.60 * 10-19 C2 11.0 * 104 m>s2 12.5 * 10-5 T2sin 90
= 4.0 * 10220 N
(b) The electron has the same magnitude of
charge as the proton and the same velocity, so
it experiences the same magnitude of magnetic
force. But its charge is negative, so the direction of the magnetic force Fs on the electron is
opposite to the direction given by the righthand rule. Hence the force Fs points vertically
downward.
Charge on the electron:
q = 2e = 21.60 * 10
219
C
B
q = –e
The electron velocity vector is
perpendicular to the magnetic
field vector (u = 90°), so the
magnitude of the magnetic
force on the electron is
East
up
v
North
F
East
F =  q  vB sin u = evB sin u
= 11.60 * 10-19 C2 11.0 * 104 m>s2 12.5 * 10-5 T2sin 90
= 4.0 * 10220 N
(c) The proton velocity s
v and the magnetic
s again both lie in a horizontal plane, but
field B
now the angle between these vectors is u = 25°.
So the magnitude of the magnetic force is less
than in part (a). The direction of the force is
the same as in part (a), however.
The proton velocity vector is at
an angle u = 25° to the magnetic
field vector, so the magnitude
of the magnetic force on the
proton is
F
q = +e
F =  q  vB sin u = evB sin u
= 11.60 * 10-19 C2 11.0 * 104 m>s2
* 12.5 * 10-5 T2 sin 25
= 1.7 * 10220 N
25°
B
v
up
North
East
This is smaller than in part (a) because sin 25° = 0.42 compared to
sin 90° = 1.
Reflect
This example illustrates how the magnetic force on a moving charged particle depends on the direction in which the
particle is moving. Note that the force magnitudes in parts (a), (b), and (c) are very small because a single electron or
proton carries very little charge. These particles also have very small mass, however (1.67 * 10227 kg for the proton and
9.11 * 10231 kg for the electron), so the resulting accelerations are tremendous. Can you use Newton’s second law to
show that the proton in part (a) and the electron in part (b) have accelerations of 2.4 * 107 m>s 2 and 4.4 * 1010 m>s 2
respectively?