Need help. Please show work, I will leave a like A 5-meter long copper wire with

Question

Need help. Please show work, I will leave a like

A 5-meter long copper wire with square cross-section has 48 c of charge pass through it in one millisecond. For this to occur, a potential difference of 1 mV (1x10 3 V) exists between the ends of the wire. Using 6x10 +7 -1m-1 for the conductivity of copper, and Ohm's Law:-oE 1. a) b) Find x, the side length of the wire in meters (see diagram for clarity). Find the resistance, R, of this wire in ohms. Length, L THE WIRE LENGTH IS NOW DOUBLED, AND IT IS RESHAPED INTO A CIRCULAR CROSS-SECTION, WITH DIAMETER X c) Find the new resistance, R', in terms of R.

Explanation / Answer

Given,

Q = 48 uC ; L = 5 m ; t = 10^-3 s ; V = 10^-3 V

We know from defination of current

I = Q/t

I = 48 x 10^-6/10^-3 = 48 mA = 0.048 A

from Ohm's law,

V = IR => R = V/I

R = 10^-3/0.048 = 0.021 Ohm

We know that resistance is given by:

R = rho L/A

A = rho L/R

x^2 = rho L/A

x = sqrt (rho L/A)

rho = 1/6 x 10^7 = 1.67 x 10^-8Ohm m

x = sqrt (1.67 x 10^-8 x 5 / 0.021) = 0.00199 m

Hence, x = 0.00199 m = 1.99 milli meter ------------> Ans(a)

b)R = 0.021 Ohm

c)L' = 10 m ;

R' = rho L'/A' = rho 10/pi 0.00199^2

R = rho 5/0.00199^2

R'/R = rho 10/pi x 0.00199^2 x 0.00199^2/5 = 0.64

R' = 0.64 R

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