Unformatted text preview: sing Equation 19.3 to express the electric potential energy of the charge q0 as EPE = q0V,
where V is the electric potential, we find from Equation (1) that
1
2
mvB
2 1
2 2
+ q0VB = mvA + q0VA (2) Since the particle starts from rest, we have that vA = 0 m/s, and Equation (2) indicates that vB = 2q0 (VA − VB )
m ( )( ) 2 2 +1.60 ×10−19 C 1.20 ×106 V =
= 1.08 ×107 m/s
−27
6.64 ×10
kg b. According to Equation 21.1, the magnitude of the magnetic force that acts on the particle
is ( )( ) F = q0 vB B sin θ = 2 1.60 × 10−19 C 1.08 × 107 m/s ( 2.20 T ) sin 90.0° = 7.60 × 10−12 N where θ = 90.0°, since the particle travels perpendicular to the field at all times.
c. According to Equation 21.2, the radius of the circular path on which the particle travels is ( 6.64 ×10−27 kg ) (1.08 ×107 m/s ) = 0.102 m
r=
=
q0 B
2 (1.60 ×10−19 C ) ( 2.20 T )
mvB 22. REASONING The radius of the circular path is given by Equation 21.2 as r = mv/( q B),
where m is the mass of the species, v is the speed, q is the magnitude of the charge, and B
is the magnitude of the magnetic field. To use this expression, we must know something
about the speed. Information about the speed can be obtained by applying the conservation
of energy principle. The electric potential energy lost as a charged particle “falls” from a
higher to a lower electric potential is gained by the particle as kinetic energy. SOLUTION For an electric potential difference V and a charge q, the electric potential
2
1
energy lost is q V, according to Equation 19.4. The kinetic energy gained is 2 mv . Thus,
energy conservation dictates that
q V = 1 mv 2
2 or v= 2qV
m Substituting this result into Equation 21.2 for the radius gives Chapter 21 Problems r= mv
m 2 qV 1
=
=
qB qB
m
B 1151 2mV
q Using e to denote the magnitude of the charge on an electron, we note that the charge for
+
2+
species X is +e, while the charge for species X is +2e. With this in mind, we find for the
ratio of the radii that
1 2 mV
r1
B
e
=
= 2 = 1.41
r2
1 2 mV
B
2e 23. SSM REASONING When the proton moves in the magnetic field, its trajectory is a
circular path. The proton will just miss the opposite plate if the distance between the plates
is equal to the radius of the path. The radius is given by Equation 21.2 as r = mv / ( q B ) . This relation can be used to find the magnitude B of the magnetic field, since values for all
the other variables are known.
SOLUTION Solving the relation r = mv / ( q B ) for the magnitude of the magnetic field, and realizing that the radius is equal to the plate separation, we find that ( )( ) 1.67 × 10−27 kg 3.5 × 106 m/s
mv
B=
=
= 0.16 T
qr
1.60 × 10−19 C ( 0.23 m ) ( ) The values for the mass and the magnitude of the charge (which is the same as that of the
electron) have been taken from the inside of the front cover. 24. REASONING The magnitude FB of the magnetic
force acting on the particle is related to its speed v
by FB = q0 v B sin θ (Equation 21.1), where B is
the magnitude of the magnetic field, q0 is the
particle’s charge, and θ is the angle between the
magnetic field B and the particle’s velocity v. As
the drawing shows, the vector v (east, to the right)
is perpendicular to the vector B (south, out of the
page). Therefore, θ = 90°, and Equation 21.1
becomes Up
B (out of page)
E West v Down East 1152 MAGNETIC FORCES AND MAGNETIC FIELDS FB = q0 v B sin 90o = q0 v B (1) In addition to the magnetic force, there is also an electric force of magnitude FE acting on
the particle. This force magnitude does not depend upon the speed v of the particle, as we
see from FE = q0 E (Equation 18.2). The particle is positively charged, so the electric force
acting on it points upward in the same direction as the electric field. By RightHand Rule
No.1, the magnetic force acting on the positively charged particle points down, and is
therefore opposite to the electric force. The net force on the particle points upward, so we
conclude that the electric force is greater than the magnetic force. Thus, the magnitude F of
the net force acting on the particle is equal to the magnitude of the electric force minus the
magnitude of the magnetic force: F = FE − FB (2) We also note that particles traveling at a speed v0 = 6.50×103 m/s experience no net force.
Therefore, FE = FB for particles moving at the speed v0.
SOLUTION Substituting Equation (1) and FE = q0 E (Equation 18.2) into Equation (2)
yields
(3)
F = q0 E − q0 v B
The magnetic field magnitude B is not given, but, as noted above, for particles with speed
v0 = 6.50×103 m/s, the magnetic force of Equation (1), FB = q0 v0 B , is equal to the electric
force FE = q0 E (Equation 18.2). Therefore, we have that
q0 v0 B = q0 E or B= E
v0 (4) Substituting Equation (4) into Equation (3) yields
F = q0 E − q0 v B = q0 E − q0 v E (5) v0 Solving Equation (5) for v, we obtain
q0 v E
v0 = q0 E − F or v
F
= 1−
v0
q0 E or Substituting the given values into Equation (6), we find that F
v = v0 1 − q0 E (6) Chapter 21 Problems 1153 F
1.90 ×10−9...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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