Chapter 38 - 38.1. Model: Current is defined as the rate at...

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38.1. Model: Current is defined as the rate at which charge flows across an area of cross section. Solve: Since the current is / and / Qt QN e ΔΔ Δ= , the number of electrons per second is 8 10 1 10 1 19 10 nA 1.0 10 C/s 6.25 10 s 6.3 10 s 1.60 10 C N te × == = × × Δ×
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38.2. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 38.7 and 38.8. Solve: (a) The speed with which a particle can pass without deflection is 3 7 3 / 600 V/5.0 10 m 6.0 10 m/s 2.0 10 T EV d v BB Δ× == = = × × (b) The radius of cyclotron motion in a magnetic field is 31 7 19 3 9.11 10 kg 6.0 10 m 0.17 m 17 cm 1.60 10 C mv r eB −− ⎛⎞ ×× ⎜⎟ ⎝⎠
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38.3. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: 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 38.4. The magnitude of the magnetic field is 3 3 6 / 200 V/8.0 10 m 5.0 10 T 5.0 10 m EV d B vv Δ× == = = × × Thus G B = (5.0 × 10 –3 T, out of page).
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38.4. Model: Assume the electric field ( E = Δ V / d ) between the plates is uniform. Visualize: Please refer to Figure 38.9. Solve: (a) The mass of the droplet is () 3 336 1 61 6 drop 44 885 kg/m 0.4 10 m 2.37 10 kg 2.4 10 kg 33 mV R ππ ρρ −− ⎛⎞ == = × = × ≈× ⎜⎟ ⎝⎠ (b) In order for the upward electric force to balance the gravitational force, the charge on the droplet must be ( )( ) 16 2 drop 18 18 drop 3 2.37 10 kg 9.8 m/s 1.28 10 C 1.3 10 C 20 V 11 10 m mg q E × = × × × (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 18 droplet 19 1.28 10 C 8 1.60 10 C q N e × = ×
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38.5. Model: Assume the electric field ( E = Δ V / d ) between the plates is uniform. Visualize: Please refer to Figure 38.9. To balance the weight, the electric force must be directed toward the upper electrode, which is more positive than the lower electrode. Solve: Since () 3 4 drop 3 mV R π ρρ == , the equation drop drop mgqE = is ( ) ( ) ( ) ( ) 19 3 19 3 32 4 4 3 3 25 V 0.012 m 15 1.60 10 C 15 1.4163 10 m 0.52 m 860 kg/m 9.8 m/s Vd e RR g μ ρ × Δ = × =
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38.6. Model: The charge on an object is an integral multiple of a certain minimum charge value. Solve: From smallest charge to the largest charge, the measured charges on the drops are 2.66 × 10 19 C, 3.99 × 10 19 C, 6.65 × 10 19 C, 9.31 × 10 19 C, and 10.64 × 10 19 C. The differences between these values are 1.33 × 10 19 C, 2.66 × 10 19 C, 2.66 × 10 19 C, and 1.33 × 10 19 C. These differences are a multiple of 1.33 × 10 19 C.
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Chapter 38 - 38.1. Model: Current is defined as the rate at...

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