Question

Problems 18, 19 and 20 pertain to the following: A straight 2.5m long silver wire of diameter 0.30mm is connected across a 3.0V potential difference. 18.) [2 points] How many electrons have passed through the wire in 5 minutes? a.) 5.8 b.) 1731 c.) 1.1x1022 d.) 3.1x1027 e.) none of these 19.) [2 points] How much energy does this wire dissipate during 5 minutes? a.) 17.3J b.) 5.2x103 c.) 3.1x1027 d.) 0J e.) none of these

Answers are C, B, A but how do you solve them?

Answer 1

The solusolution is given below

Given, e-2.5m d = 0.30mm = 3x10-4 m So, - d = 2 x 10-am = 1.5x104m. Area, A - Thr2 = 3.14 x(1.5 x10-432 >> A = 7:065X10-8m2 V = 3.0v. Also, for silver, resistivity p = 159 x10-82m So, resistance R = pl.l.59x10-8 x 2.5 7.065810-8 - R = 0.56 18) Now, I= - 3.0 = 5.4A 0.56 We know that, I= = ne n = It Given, t= 5 minutes 5860 = 300 sec. = n = 5.4 x 300 1.6 X 10-19 - n = 1.03x1022 electrons. The answer can be opted as (C) 19) Energy, E=I2RE E = (5.4) ² x 0.56 x 300 E = 5 x 1031 Here, the answer will be (b)

20) volume change density, P a) l= Int A opv Given 8 - 5.8 X 1028 electrons per nebic meter. = 5.8 x 1028 × 1.6x1049 cm 3 – 9.28x109 m 3 t - 2 minute = 60 sec. =) ( = 5.4 X 60 7.065X10-8 X 9.28 X10-9 ->l = 0.5ml The connect option is ca).