Unformatted text preview: Physics 204, morning summer 2008 Homework 6 solutions Chapter 21 Problems 5,9,14,15,17 5. SSM REASONING AND SOLUTION The magnitude of the force can be determined using Equation 21.1, F = q vB sin where is the angle between the velocity and the magnetic field. The direction of the force is determined by using Right‐Hand Rule No. 1. a. F = q vB sin 30.0° = (8.4 10 –6 C)(45 m/s)(0.30 T) sin 30.0° = 5.7 × 10 −5 N , directed into the paper . b. F = q vB sin 90.0° = (8.4 10 –6 C)(45 m/s)(0.30 T) sin 90.0° = 1.1 × 10 −4 N , directed into the paper . c. F = q vB sin 150° = (8.4 10 –6 C)(45 m/s)(0.30 T) sin 150° = 5.7 × 10 −5 N , directed into the paper . 9. SSM WWW REASONING The direction in which the electrons are deflected can be determined using Right‐
Hand Rule No. 1 and reversing the direction of the force (RHR‐1 applies to positive charges, and electrons are negatively charged). Each electron experiences an acceleration a given by Newton’s second law of motion, 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 , so it is the net force. The speed v of the electron is related to its kinetic energy KE 1
by the relation KE = 2 mv 2 . Thus, we have enough information to find the acceleration. SOLUTION a. According to RHR‐1, 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, a = q vB sin θ
F
= 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 a = = q0 vB sin θ = m 2(KE)
B sin θ
m
m q0 (1.60 × 10−19 C) ( 2 2.40 × 10
9.11 × 10 −15 −31 J ) kg 9.11 × 10 −31 ( 2.00 × 10−5 T ) sin 90.0° kg = 2.55 × 1014 m/s 2 14. REASONING AND SOLUTION The radius of the circular path of a charged particle in a magnetic field is given by ⎛
⎜
⎝ Equation 21.2 ⎜ r = mv ⎞
⎟ . 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 . 15. SSM REASONING AND SOLUTION The radius of curvature for a particle in a mass spectrometer is discussed in Section 21.4. According that discussion, the radius for a charged particle with a charge of +e (e = 1.60 × 10−19 C) is r= 2mV
2mV
. In this problem, the charged particle has a charge of +2e, so that the radius becomes r =
. 2
eB
2eB 2 Thus, the desired radius is r= 2mV
=
2eB 2 ( )(
)
2
2 (1.60 × 10−19 C ) ( 0.500 T ) 2 3.27 × 10−25 kg 1.00 × 103 V = 0.0904 m 17. REASONING The drawing shows the velocity v of the as they enter the magnetic field B. The diameter of the followed by the carbon‐12 atoms is labeled as 2r12, and B (out of paper) carbon atoms circular path that of the carbon‐13 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 charge on the ion difference d in the diameters is Δd = 2r13 − 2r (see the 12 SOLUTION The spatial separation between the two they have traveled though a half‐circle is v q 2r12
2r13 is the (q = +e). The drawing). isotopes after ⎛ m v ⎞ ⎛ m v ⎞ 2v
Δd = 2r13 − 2r12 = 2 ⎜ 13 ⎟ − 2 ⎜ 12 ⎟ =
( m13 − m12 )
⎝ eB ⎠ ⎝ eB ⎠ eB
= ( 2 6.667 × 10 m/s (1.60 × 10 5 −19 ) ( 21.59 ×10−27 kg − 19.93 ×10−27 kg ) = 1.63 × 10−2 m
C ) ( 0.8500 T ) ...
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 Spring '09
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 Physics, Tour de Georgia, ΔD, vB sin

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