What is the magnitude of drift velocity of the conduction electron? Answer 6.71

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What is the magnitude of drift velocity of the conduction electron? Answer 6.71 Times 10^(-4) m/s. 5.31 Times 10^(-1) m/s. 5.31 Times 10^(-4) m/s^2. 6.71 Times 10^(-1) m/s^2. 3.80 Times 10^(1) m/s. What is the magnitude of the electric field in the block? Answer: 1.26 C/N 7.19 C/N 1.38 N/C 5.38 V/m 1.19 V/m In the lower atmosphere of the Earth there are negative and positive ions, created by indicated electron in the soil and cosmic rays from space. Suppose that at a certain time in certain region of the atmosphere, the atmospheric electric field strength is 1120 V/m, directed vertically down. Due to this field, singly charged positive ions, 7600 cm^2, drift downward and singly charged negative ions, 2150/cm^3 upward. The measured conductivity is 5.70 Times 10^-14 (ohm-m)^-1. Calculate the ion drift signal. Assuming the same speed for positive and negative ions. Answer: 4.58 Times 10^(-2) m/s 7.3 Times 10^(-1) m/s 5.91 Times 10^(-4) m/s 5.08 Times 10^(-2) m/s 8.80 Times 10^(1) m/s Assuming that the potential stays constant across the terminals of a 12.0 V battery that has an initial charger of 91 AA how long (in hours) will it be able to deliver energy at a rate of 45.0 W? Answer: 2,97 h. 24.3 h. 6.54 h. 26.1 h. 3.87 h. In figure 5 of Assignment 7, epsilon = 21.0 kV, C = 3800 pF, R_1 = R_2 = R_3 = 6.80 M Ohm. At time t - 0 (i.e. immediately after the circuit is connected) the current thought R_1 = Answer: 2.97 mA. 2.44 A. 6.54 mA. 1.97 mA. 3.87 mA. At t = 0 (i.e. immediately after the circuit is connected) the current thought R_2 = Answer: 1.97 mA. 1.44 A. 6.00 mA. 4.97 mA. 8.97 mA At t = 0 (i.e. immediately after the circuit is connected) the current throught R_3 = Answer: 1.97 mA. 1.44 A. 6.00 mA. 4.97 mA. 8.97 mA What is the current in R_3 at t = infinity ? Answer: 9.97 mA. 2.04 A. 1.54 mA. 2.97 mA. 2.87 mA. What is the potential difference across R_1 at t = infinity? Answer: 1.97kV. 8.44 V. 3.54 kV. 9.97 V. 9.11 kV. What is the potential difference across R_2 a very long time after the circuit is connected? Answer: 1.97kV. 8.44 V. 3.54 kV. 9.97 V. 9.11 kV. A proton moves in a circle of radius 1.26 m in a region where the magnetic field is 99.0 mT. Determine the frequency of the oscillations. Answer: 2.97 Hz. 1.37 MHz. 6.54Hz. 1.97 MHz. 3.87 MHz. Determine the kinetic energy, in millions electron volts, i.e., MeV of the proton. (Ignore any relativistic effects) Answer: 2.97 MeV. 2.44 eV. 6.54 eV. 1.97 MeV. 0.616 MeV. See figure 2 of Assignment 8. A rod of mass 210 g and 75.0 cm long is suspended by two springs ina 3.50 T magnetic field, as shown. The weight of the rod causes the springs to be in tension and they sketch a little. A current is now set up in the rod and the magnitude and direction of the current is such that weight of the rod is now offset and the tension in the scorings is released. Determine the magnitude of the current in the rod. Answer: 2.97 MeV. 2.44 eV. 6.54 mA. 1.97 mA. 3.87 mA. The direction of the rod is from Answer: Left to Right. Right ti Left.

Explanation / Answer

10)

IN the magetic field

Fc = Fb

mv^2/r = qvB

r = mv/qB

timeperiod T = 2pir/v = 2pim/qB

frequency f = 1/T = qB/(2*pi*m) =

f = (1.609*10^-19*99*10^-3)/(2*pi*1.67*10^-27)

f = 1.52 MHz

kinetic energy

K = 0.5*m*v^2

r = mv/qB

v = rqB/m

k = 0.5*r^2*q^2*B^2/m

k = 0.5*1.26^2*(1.6*10^-19)^2*0.099^2/(1.67*10^-27)

k = 1.19*10^-13 J

k = 1.19*10^-13/(1.6*10^-19)

k = 0.74 MeV

+++++++++++++++++++

In equilibirium

Fb = Fg

I*L*B = m*g

I*0.75*3.5 = 0.21*9.8

I = 0.784 A

directionright to left

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