Unformatted text preview: the proton and the electron, both of which carry
charges of the same magnitude q = e, we obtain r= mp v eBp
1 24
43 mv
r= e
eBe
1 24
43 and Electron Proton Dividing the protonequation by the electronequation gives mp v
r eBp
=
r me v
eBe 1= or mp Be
me Bp Solving for Be, we obtain Be = me Bp
mp (9.11×10−31 kg ) ( 0.50 T ) =
=
1.67 ×10 −27 kg 2.7 ×10−4 T Chapter 21 Problems 1147 17. SSM REASONING AND SOLUTION The radius of the circular path of a charged mv particle in a magnetic field is given by Equation 21.2 r =
. q B a. For an electron ( )(
)( ) 9.11 × 10−31 kg 9.0 × 106 m/s
mv
r=
=
=
qB
1.6 × 10−19 C 1.2 × 10−7 T ( ) 4.3 × 102 m b. For a proton, only the mass changes in the calculation above. Using m = 1.67 × 10 –27 kg, we obtain r = 7.8 × 105 m . 18. REASONING As discussed in Section 21.4, the mass m of a singlyionized particle that has
been accelerated through a potential difference V and injected into a magnetic field of
magnitude B is given by er 2 2
(1)
m= 2V B where e = 1.60 × 10 −19 C is the magnitude of the charge of an electron and r is the radius of
the particle’s path. If the beryllium10 ions reach the same position in the detector as the
beryllium7 ions, both types of ions must have the same path radius r. Additionally, the er 2 accelerating potential difference V is kept constant, so we see that the quantity in 2V Equation (1) is the same for both types of ions. SOLUTION All that differs between the two situations are the masses (m7, m10) of the ions
and the magnitudes of the magnetic fields (B7, B10). Solving Equation (1) for the constant er 2 quantity , we obtain 2V er 2 m10 m7 2V = B 2 = B 2 2
10
7
13
Same for
both ions 2
Solving Equation (2) for B10 , we find that (2) 1148 MAGNETIC FORCES AND MAGNETIC FIELDS 2
2
B10 = B7 m10
m7 B10 = B7 or m10
m7 = ( 0.283 T ) 19. REASONING The drawing shows the velocity v of
the carbon atoms as they enter the magnetic field B.
The diameter of the circular path followed by the
carbon12 atoms is labeled as 2r12, and that of the
carbon13 atoms as 2r13, where r denotes the radius of
the path. The radius is given by Equation 21.2 as
r = mv / ( q B ) , where q is the charge on the ion (q = +e). The difference ∆d in the diameters is
∆d = 2r13 − 2r12 (see the drawing). 16.63 × 10−27 kg
= 0.338 T
11.65 ×10−27 kg B (out of paper) v 2r12
2r13 SOLUTION The spatial separation between the two isotopes after they have traveled
though a halfcircle is m v m v 2v
∆d = 2r13 − 2r12 = 2 13 − 2 12 =
( m13 − m12 ) eB eB eB
= ( 2 6.667 ×105 m/s (1.60 × 10 −19 ) ( 21.59 ×10−27 kg − 19.93 ×10−27 kg ) = 1.63 × 10−2 m
C ) ( 0.8500 T ) 20. 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 can determine the speed of each particle by employing the principle of conservation of
energy. The electric potential energy lost as the particles accelerate is converted into kinetic
energy. Equation 19.4 indicates that the electric potential energy lost is q V, where q is
the magnitude of the charge and V is the electric potential difference. Since q and V are the
same for each particle, each loses the same amount of potential energy. Energy
conservation, then, dictates that each gains the same amount of kinetic energy. Since each
particle starts from rest, each enters the magnetic field with the same amount of kinetic
energy. SOLUTION According to Equation 21.2, r = mv/( q B). To determine the speed v with
which each particle enters the field, we use Equation 19.4 and the energyconservation
principle as follows: Chapter 21 Problems = qV
{
Electric potential
energy lost 1 mv 2
22
13 v= or 1149 2 qV
m Kinetic energy
gained Substituting this result into Equation 21.2 gives the radius of the circular motion: r= mv
m 2 qV 1
=
=
qB qB
m
B 2mV
q Applying this result to each particle, we obtain 1 2m1V
r1 =
B
q
14 244
4
3 1 2m2V
r2 =
B
q
14 244
4
3 and Particle 1 Particle 2 Dividing r2 by r1 gives
1 2m2V
B
q r2
=
=
r1
1 2m1V
B
q r2 = r1 m2
m1 m2
5.9 × 10−8 kg
= (12 cm )
= 19 cm
m1
2.3 × 10−8 kg 21. REASONING The speed of the αparticle can be obtained by applying the principle of
conservation of energy, recognizing that the total energy is the sum of the particle’s kinetic
energy and electric potential energy, the gravitational potential energy being negligible in
comparison. Once the speed is known, Equation 21.1 can be used to obtain the magnitude of
the magnetic force that acts on the particle. Lastly, the radius of its circular path can be
obtained directly from Equation 21.2. SOLUTION
a. Using A and B to denote the initial positions, respectively, the principle of conservation
of energy can be written as follows:
1
mv 2
2
1 24
4B
3
Final kinetic
energy + 1 EPE
=
mv 2 + EPE A
2
1 24
4B
3
1 24
43
1 24
4A
3 Final electric
potential energy Initial kinetic
energy Initial electric
potential energy (1) 1150 MAGNETIC FORCES AND MAGNETIC FIELDS U...
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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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