Chapter37 - 37.1. Visualize: By showing that a current is...

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37.1. Visualize: By showing that a current is collected only when the cathode-ray green spot is on the electrode, this was the first conclusive demonstration that cathode rays consist of negatively charged particles.
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37.2. Visualize: Cathode rays emitted by various hot cathodes have the same charge-to-mass ratio independent of the material used in the hot cathode.
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37.3. Model: Current is defined as the rate at which charge flows across an area of cross section. Solve: Since the current is ∆∆ Qt QN e and = , the number of electrons per second is N te == × × 10 nA C / s C s 10 10 160 10 625 10 8 19 10 1 . . .
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37.4. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 37.7 and 37.8. Solve: (a) The speed with which a particle can pass without deflection is v E B Vd B == = × × 600 V m T ±m/s 50 10 20 10 60 10 3 3 7 . . . (b) The radius of cyclotron motion in a magnetic field is r m e v B = = × × × × −− 911 10 160 10 31 19 7 3 . . . . kg C m T 0.171 m 17.1 cm
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37.5. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figure Ex37.5. Without the external magnetic field B , the electrons will be deflected up toward the positive electrode. The magnetic field must therefore be directed out of the page to exert a balancing downward force on the negative electron. Solve: In a crossed-field experiment, the magnitudes of the electric and magnetic forces on the electron are given by Equation 37.4. The magnitude of the magnetic field is B E v Vd v == = × × 200 V m m T 80 10 50 10 3 6 3 . . . Thus r B = (5.0 × 10 –3 T, out of page).
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37.6. Model: Assume the electric field ( E = V / d ) between the plates is uniform. Visualize: Please refer to Figure 37.9. Solve: (a) The mass of the droplet is mV R drop 3 885 kg / m m kg == = () × −− ρρ ππ 4 3 4 3 04 10 237 10 36 3 16 .. (b) In order for the upward electric force to balance the gravitational force, the charge on the droplet must be q mg E drop drop 2 kg 9.8 m / s 20 V m C × × 11 10 128 10 16 3 18 . . (c) Because the electric force is directed toward the electrode at the higher potential (or more positive plate), the charge on the droplet is negative. The number of surplus electrons is N q e × × = droplet C C 160 10 8 18 19 . .
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37.7. Model: Assume the electric field ( E = V / d ) between the plates is uniform. Visualize: Please refer to Figure 37.9. To balance the weight, the electric force must be directed toward the upper electrode, which is more positive than the lower electrode. Solve: Since mV R drop ==() ρρ π 4 3 3 , the equation mgqE drop drop = is R Vd e g R 3 4 3 19 4 3 19 15 1 60 10 1 4163 10 0 521 = () = () × ( ) = 15 25 V 0.012 m C 860 kg / m 9.8 m / s m m 32 3 ρ µ .
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This note was uploaded on 02/27/2010 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter37 - 37.1. Visualize: By showing that a current is...

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