the ones in red are the answers, but i need the steps Problem 10: Two metallic w

Question

the ones in red are the answers, but i need the steps

Problem 10: Two metallic wires of gold have the same length of 2.0 m but different cross-sectional area. Diameter of wire A is 2 mm while diamctcr of wire B is 1 mm. The two wires are connected in series and then to a (10 V) battery as shown in the figure. wire wire A B 10V Assume the resistivity of gold is 2.44x10 52.m. Number density of free electrons in gold is n 5.90 x 102 m (a) What is the resistance (R) of each wire? (b) What is the current through cach wire? (c) What is the current density U) in each wire? (d) What is the drift velocity of the electrons inside each wire? (e) How much time, on average, is required for electrons leaving the negative terminal of the battery to reach the positive terminal? c) 1.6x10 Am a R1 0.0162 R2 0.0622 b 128.2A d) A 4.33x10 m/s CVd BJ 1.73x10 m/s e) 5.75x10 s

Explanation / Answer

(a)

Resistivity = (R*A)/l

Cross-sectional area AA = pi*r^2 = 3.142 * (2*10^-3)^2 = 1.025 *10^-5m sq

Cross-sectional area AB = pi*r^2 = 3.142 * (1*10^-3)^2 = 3.142 *10^-6m sq

RA = (resistivity*l)/A =(2.44*10^-8 * 2)/ 1.025 *10^-5 = 4.76m

RB = (resistivity*l)/A = (2.44*10^-8 * 2)/ 3.142 *10^-6 = 15m

(b)

IA = V/R = 10/4.76*10^-3 = 2100.8A

IB = V/R = 10/3.142*10^-3 = 666.7A

c)

Current density

JA = IA/A = 2100.8/1.025*10^-5 = 204.96 MA/m sq.

JB = IB/A = 666.7/3.142*10^-6 = 212.2 MA/m sq.

d)

Drift velocity

uA = IA/nAq where n = 5.90*10^28 m^-3, q=-1.6*10^-19 C

= 2100.8/(5.90*10^28 m^-3 * -1.6*10^-19 C * 1.025 *10^-5 = 0.022 m/s

UB = IB/nAq

= 666.7/(5.90*10^28 m^-3 * -1.6*10^-19 C * 3.142 *10^-6 = 0.071 m/s

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