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Chapter37

# 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.
37.3. Model: Current is defined as the rate at which charge flows across an area of cross section. Solve: Since the current is Q t Q N e and = , the number of electrons per second is N t e = = × × = × 10 nA C / s C s 1 0 10 1 60 10 6 25 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 V d B = = = × × = × 600 V m T m / s 5 0 10 2 0 10 6 0 10 3 3 7 . . . (b) The radius of cyclotron motion in a magnetic field is r m e v B = = × × × × = = 9 11 10 1 60 10 6 0 10 2 0 10 31 19 7 3 . . . . kg C m T 0.171 m 17.1 cm
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 V d v = = = × × = × 200 V m m T 8 0 10 5 0 10 5 0 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 m V R drop 3 885 kg / m m kg = = = ( ) × ( ) = × ρ ρ π π 4 3 4 3 0 4 10 2 37 10 3 6 3 16 . . (b) In order for the upward electric force to balance the gravitational force, the charge on the droplet must be q m g E drop drop 2 kg 9.8 m / s 20 V m C = = × ( ) ( ) × = × 2 37 10 11 10 1 28 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 1 28 10 1 60 10 8 18 19 . .
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 m V R drop = = ( ) ρ ρ π 4 3 3 , the equation m g q E drop drop = is R V d 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 3 2 3 π π ρ µ .

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

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