A 3.00 m length of copper wire at 20°C has a 1.20 m long section with diameter 1

Question

A 3.00 m length of copper wire at 20°C has a 1.20 m long section with diameter 1.60 mm and a 1.80 m long section with diameter 0.80 mm. There is a current of 2.4 mA in the 1.60 mm diameter section (a) What is the current n the 0.80 mm diameter section? 2.4 mA (b) What is the magnitude of E in the 1.60 mm diameter section? 199e-5 V/m (c) What is the magnitude of E in the 0.80 mm diameter section? 7.97e-5 V/m (d) What is the potential difference between the ends of the 3.00 m length of wire? 99e-4

Explanation / Answer

Given

total length of the copper wire is 3 m

of 1.2 m with diameter d1 = 1.6mm and 1.8m wire with diameter d2 = 0 .8 mm

total current is I = 2.4 m in 1. mm diameter wire section

we know that these wires are connected in series so the current will be same on each wier so  

first we calculate the resistance of the two sections later using Ohm's law we get the potential difference

resistivity of copper is rho_c = 1.72*10^-8 ohm m

area of cross sections are A1 = pi(d1/2)^2 = pi((1.6*10^-3)/2)^2= 2.011*10^-6 m^2

and area of cross sections A2 = pi(d2/2)^2 = pi((0.8*10^-3)/2)^2= 5.027*10^-7 m^2

now the restances are R1 = rho*l1/A1 = 1.72*10^-8*1.2/(2.011*10^-6) ohm = 0.0102635504724 ohm

R2 = rho*l2/A2 = 1.72*10^-8*1.8/(5.027*10^-7) ohm = 0.0615874278894 ohm

the total resistance in series is R = R1+R2 = 0.0102635504724+0.0615874278894 ohm = 0.0718509783618 ohm

From Ohm's law V = I*R

V = 2.4*10^-3*0.0718509783618 V = 0.0001724423481 V = 0.0001724423481 V = 1.7244*10^-4 V

the potential difference between the ends of the rod is V = 1.7244*10^-4 V

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