The energy gained per cycle is 2 q δ v 2 e 500 v j

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The proton accelerates through a potential difference of 500 V twice during one revolution. The energy gained per cycle is 2 q Δ V = 2 e (500 V) = 1.60 × 10 16 J Using the maximum kinetic energy of the proton from part (a), the number of cycles before the proton attains this energy is 13 16 4.6 10 J 2850 1.60 10 J × = ×
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33.64. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Visualize: Please refer to Figure P33.64. Solve: The potential difference causes an ion of mass m to accelerate from rest to a speed v . Upon entering the magnetic field, the ion follows a circular trajectory with cyclotron radius r = mv / eB . To be detected, an ion’s trajectory must have radius d = 2 r = 8 cm. This means the ion needs the speed 2 eBr eBd v m m = = This speed was acquired by accelerating from potential V to potential 0. We can use the conservation of energy equation to find the voltage that will accelerate the ion: K 1 + U 1 = K 2 + U 2 2 1 2 0 J 0 J e V mv + Δ = + 2 2 mv V e Δ = Using the above expression for v , the voltage that causes an ion to be detected is 2 2 2 2 2 2 2 8 mv m eBd eB d V e e m m Δ = = = An ion’s mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of 2 N + is m = m N + m N m elec = 2(14.0031 u)(1.661 × 10 27 kg/u) – 9.11 × 10 31 kg = 4.65174 × 10 26 kg Note that we’re given the atomic masses very accurately in Exercise 28. We need to retain this accuracy to tell the difference between 2 N + and CO + . The voltage for + 2 N is ( ) ( ) ( ) ( ) 2 2 19 26 1.6 10 C 0.200 T 0.08 m 110.07 V 8 4.65174 10 kg V × Δ = = × Ion Mass (kg) Accelerating voltage (V) 2 N + 4.65174 × 10 26 110.07 2 O + 5.31341 × 10 26 96.36 CO + 4.64986 × 10 26 110.11 Assess: The difference between 2 N + and CO + is not large but is easily detectable.
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33.65. Model: Assume that the magnetic field is uniform over the Hall probe. Solve: Equation 33.24 gives the Hall voltage and Equation 33.20 gives the cyclotron frequency in terms of the magnetic field. We have H IB V tne Δ = cyl 2 mf B e π = ( )( )( ) ( )( ) 27 6 3 cyc 19 3 H 2 1.67 10 kg 10.0 10 Hz 0.150 10 A 2 0.1812 T A/V 1.60 10 C 0.543 10 V mf I tne e V π π × × × = = = Δ × × With this value of tne , we can once again use the Hall voltage equation to find the magnetic field: ( ) 3 H 3 1.735 10 V 0.1812 TA/V 0.150 10 A V B tne I Δ × = = × = 2.10 T
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33.66. Model: The loop will not rotate about the axle if the torque due to the magnetic force on the loop balances the torque of the weight. Visualize: Please refer to Figure P33.66. Solve: The rotational equilibrium condition net 0 N m τ Σ = G is about the axle and means that the torque from the weight is equal and opposite to the torque from the magnetic force. We have ( ) ( ) ( ) ( )( ) ( ) ( )( )( )( ) 3 3 2 50 10 kg 0.025 m sin90 50 10 kg 9.8 m/s 0.025 m 0.123 T 10 2.0 A 0.050 m 0.100 m g B NIA B B μ × = ° = × = = Assess: The current in the loop must be clockwise for the two torques to be equal.
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33.67.
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