c According to Equation 212 the radius of the circular path on which the

C according to equation 212 the radius of the

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c. According to Equation 21.2, the radius of the circular path on which the particle travels is ( 29 ( 29 ( 29 ( 29 27 7 B 19 0 6.64 10 kg 1.08 10 m/s 0.102 m 2 1.60 10 C 2.20 T mv r q B - - × × = = = × 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 energy lost is q V , according to Equation 19.4. The kinetic energy gained is 1 2 mv 2 . Thus, energy conservation dictates that 2 1 2 2 or q V q V mv v m = = Substituting this result into Equation 21.2 for the radius gives
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127 MAGNETIC FORCES AND MAGNETIC FIELDS 2 1 2 q V mv m mV r q B q B m B q = = = Using e to denote the magnitude of the charge on an electron, we note that the charge for species X + is + e , while the charge for species X 2+ is +2 e . With this in mind, we find for the ratio of the radii that r r B mV e B mV e 1 2 1 2 1 2 2 2 1 41 = = = . 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 ( 29 / 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 ( 29 / 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 ( 29 ( 29 ( 29 ( 29 27 6 19 1.67 10 kg 3.5 10 m/s 0.16 T 1.60 10 C 0.23 m mv B q r - - × × = = = × 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 F B of the magnetic force acting on the particle is related to its speed v by B 0 sin F q v B θ = (Equation 21.1), where B is the magnitude of the magnetic field, q 0 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 Down East West B (out of page) E v
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Chapter 21 Problems 128 B 0 0 sin90 F q v B q v B = = o (1) In addition to the magnetic force, there is also an electric force of magnitude F E acting on the particle. This force magnitude does not depend upon the speed v of the particle, as we see from 0 E q F 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 Right-Hand 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
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