is the magnitude of drift velocity of the conduction clectrons? Cns7x10-5)ms. (e

Question

is the magnitude of drift velocity of the conduction clectrons? Cns7x10-5)ms. (e)s (a)6.71x104) m/s. (b)5.31x10-1) m's (c)5.31x10°(4) What is the magnitude of the electric field in the block? (5 points) Answer (a) 1.26 CN (b)7.19 C/N. (c) 1.38 N/C. (d) 538V/m. (e) 1.19V/m. ( 7) (8 points): In the lower atmosphere of the Eart : In the lower atmosphere of the Earth there are negative h there are negative and positive ions, created by radioactive elements in the soil and certain time in a certain region of the atmosphere, the atmospheric electrie field strength is r he mesured conductivity is 5.70x 10 (ohms-m)" Calculate the ion drift speed, assurming (d)6.11x10"(-5) cosmic rays from space. Suppose that at a certain time in a certain region of the atmosphere, the drift downward and singly charged 1120 V/m, directed vertically down. Due to this field, singly charged positive ions negative ions, 2650ens. drift upward. The measured conductivity is 5-70 x10" the same speed for positive and negative ions Answer: (a)4.58x10%23mt/s. (b)731x10°(-17m/s. (c)5 91 x10'd-4)m/s. potential stays constant across the terminals of a 12.0 V battery that has an initial charge of 98 A.h s) (6 points): Assuming that the potential stays constant how long (in hours) will it be able to deliver energy at a rate of 45.0 W Answer: (a) 2.97 h. (b)2.44 h. (c)6.54 h. (d) 26.lh. (e)3.87h( Attime t -o (Le, immediately 9)20 points): In Figure S of Assignment 7, g-25.0 kV. C-3800 pF. R, -R, -R,-680 M 9) (20 points): In Figure S of Assignment 7,S-25.0 kV.c-3 circuit is connected) the current through R after the Answer: (a)2.97 mA. (b)2.44 A. ()6.54 mA. (d) 1.97 mA (e)3.87 mA (_ Att-0(ie, immediately after the circuit is connected) the current in R Answer: (a) 1.97 mA. (b) 1.44 A. (c) 6.00 (d) 4.97 mA. (e) 8.97mA. (f)- At t-0 (ie, immediately after the circuit is connected) the current in R Answer: (a) L97 mA· (b) 1.44 A. (c) 6.00 mA. (d) 4 97 mA. (e) 8.97 m1A. ( Ra, What is the current in RJ at- ? Answer: (a) 9.97 mA. (b) 2.04 A. (c) 1.54 mA. (d) 2.97 mA. (e)2.87 mA. ( what is the potential difference across R1 t ? Answer: (a) 1.97kV. (b) 8.44 V. (c)3.54 kV. (d)9.97 V. (e) 9.11kv. (0 What is the potential difference across R a very long time after the circuit is connected? Answer (a) I.97kV. (b) 8.44 V. (c) 3.54 kV. ()9.97 V. (e) 9.1Ikv. ( 10) (10 points): A proton moves in a circle of radius 1.26 m in a region where the magnetic field is 90.0 mT. Determine the frequen of the oscillations. Answer: (a) 297 (c) 6.54 Hz. (d) L97 MIlz (e) 3.87MHz. (f)- (b) 1.37MHz. Determine the kinetic energy, in millions electron volts, i.e., MeV, of the proton. (Ignore any relativistic effects.) Answer: (a) 2.97 MeV. (b) 2.44 eV. ()6.54 eV. (d) 1.97 MeV. (e)0.616MeV. 1I) (12 points): See Fig. 2 of Assignment 8. A rod of mass 250 g and 75.0 cm long is suspended by two springs in a 3.50 T magr field, as shown. The weight of the rod causes the springs to be in tension and they stretch a little. A current is now set up in the r the magnitude and direction of the current is such that weight of the rod is now offiet and the tension in the springs is relaxed. Determine the magnitude of the current in the rod Answer: (a) 2.97 mA. (b) 2.44 A. (c) 6.54 mA. (d) 97 mA. (e) 3.87m1A. The direction ofthe rod is from Answer: (a) Left to Right. (b)Right to Left

Explanation / Answer

8 . V = 12 volt

energy of battery = 98 A h

Power = 45 W

Power = energy / time

45 = 98 / t

t = 2.18 hrs

9.

at t = 0 just after

(PD cross capacitor can not change instantly )

hence Vc = 0

R2 and R3 are in parallel.

1/R' = 1/R2 + 1/R3 = 1/6.80 + 1/6.80

R' = 3.4 M ohm

now R' and R are in series.

Req = R1 + R' = 6.8 +3.4 = 10.2 M ohm

I1 = V/Req = (25 x 10^3 Volt) / (10.2 x 10^6) = 2.44 x 10^-3 A = 2.44 mA

Ans(B)

now current will get divided into r2 and R3.

I2 = I3 = 2.44/2 = 1.22 m A ......Ans(F) in both

after long time, capacitor will fully charged hence no current will flow through R2 and C.

i2 = 0

i1 = i3 = (25 x 10^3 ) / (13.6 x 10^6) = 1.84 mA

potential diff across R3 = 1.84 x 10^-3 x 6.80 x 10^6 = 12.5 kV

PD across R1 = 12.5 kV

PD across R2 = 0

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