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Unformatted text preview: a = F/m, where F is the net force and m is the mass of the electron. The only force acting on
the electron is the magnetic force, F = q0 vB sin θ, so it is the net force. The speed v of the
electron is related to its kinetic energy KE by the relation KE =
enough information to find the acceleration. 1
2 mv 2 . Thus, we have SOLUTION
a. According to RHR1, if you extend your right hand so that your fingers point along the
direction of the magnetic field B and your thumb points in the direction of the velocity v of
a positive charge, your palm will face in the direction of the force F on the positive charge.
For the electron in question, the fingers of the right hand should be oriented downward
(direction of B) with the thumb pointing to the east (direction of v). The palm of the right
hand points due north (the direction of F on a positive charge). Since the electron is
negatively charged, it will be deflected due south .
b. The acceleration of an electron is given by Newton’s second law, where the net force is
the magnetic force. Thus,
q vB sin θ
F
a=
=0
m
m
1
Since the kinetic energy is KE = 2 mv 2 , the speed of the electron is v = 2 ( KE ) / m . Thus,
the acceleration of the electron is 1144 MAGNETIC FORCES AND MAGNETIC FIELDS a= = q0 vB sin θ = m (1.60 × 10 −19 C 2(KE)
B sin θ
m
m q0 ) ( 2 2.40 × 10−15 J
9.11 × 10 −31 ) kg 9.11 × 10 −31 ( 2.00 × 10−5 T ) sin 90.0° kg 12. REASONING The radius r of the circular path is given by r = = 2.55 × 1014 m/s 2 mv
(Equation 21.2), where
qB m and v are the mass and speed of the particle, respectively, q is the magnitude of the
charge, and B is the magnitude of the magnetic field. This expression can be solved directly
for B, since r, m, and v are given and q = +e, where e = 1.60 × 10−19 C.
SOLUTION Solving Equation 21.2 for B gives ( )( ) 3.06 × 10−25 kg 7.2 × 103 m/s
mv
B=
=
= 0.14 T
qr
+1.60 × 10−19 C ( 0.10 m ) 13. SSM REASONING AND SOLUTION mv a. The speed of a proton can be found from Equation 21.2 r =
, q B v= q Br
m = (1.6 ×10 –19 C)(0.30 T)(0.25 m)
= 7.2 ×106 m/s
–27
1.67 × 10
kg b. The magnitude Fc of the centripetal force is given by Equation 5.3, Fc = mv 2 (1.67 ×10 –27 kg)(7.2 × 106 m/s) 2
=
= 3.5 ×10 –13 N
r
0.25 m Chapter 21 Problems 1145 14. REASONING The time t that it takes the particle to complete one revolution is the time to
travel a distance d = 2π r equal to the circumference of a circle of radius r at a speed v. From
d Equation 2.1, we know that speed is the ratio of distance to elapsed time v = , so the
t elapsed time is the ratio of distance to speed:
t= d 2π r
=
v
v (1) Because the particle follows a circular path that is perpendicular to the external magnetic
mv
(Equation 21.2), where m is
field of magnitude B, the radius of the path is given by r =
qB
the mass and q is the magnitude of the charge of the particle. We will use Equation 21.2 to
determine the speed of the particle, and then Equation (1) to find the time for one complete
revolution.
SOLUTION Solving r = mv
(Equation 21.2) for v yields
qB
v= q Br
m = q
m Br (2) In the last step of Equation (2), we have expressed the speed v explicitly in terms of the
chargetomass ratio q/m of the particle. Substituting Equation (2) into Equation (1), we
obtain
2π r
2π r
2π
2π
t=
=
=
=
= 1.5 × 10 −8 s
8
q
q
v
( 5.7 × 10 C/kg ) ( 0.72 T )
m Br m B 15. REASONING
a. The drawing shows the velocity v of the particle at the top
of its path. The magnetic force F, which provides the
centripetal force, must be directed toward the center of the
circular path. Since the directions of v, F, and B are known,
we can use RightHand Rule No. 1 (RHR1) to determine if
the charge is positive or negative. B (out of paper)
v
F b. The radius of the circular path followed by a charged
particle is given by Equation 21.2 as r = mv / q B . The mass
m of the particle can be obtained directly from this relation, since all other variables are
known. 1146 MAGNETIC FORCES AND MAGNETIC FIELDS SOLUTION
a. If the particle were positively charged, an application of RHR1 would show that the
force would be directed straight up, opposite to that shown in the drawing. Thus, the charge
on the particle must be negative .
b. Solving Equation 21.2 for the mass of the particle gives
m= q Br
v (8.2 ×10−4 C) ( 0.48 T )(960 m ) = 2.7 ×10−3 kg
=
140 m/s 16. REASONING Equation 21.2 gives the radius r of the circular path as r = mv/( q B), where m, v, and q are, respectively, the mass, speed, and charge magnitude of the particle, and B
is the magnitude of the magnetic field. We wish the radius to be the same for both the
proton and the electron. The speed v and the charge magnitude q are the same for the
proton and the electron, but the mass of the electron is 9.11 × 10 kg, while that of the
proton is 1.67 × 10–27 kg. Therefore, to offset the effect of the smaller electron mass m in
Equation 21.2, the magnitude B of the field must be reduced for the electron.
–31 SOLUTION Applying Equation 21.2 to...
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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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