Question

using only 3 sig figs please

A wire 3.60 m long and 8.20 mm in diameter has a resistance of 15.2 mΩ. A potential difference of 22.1 V is applied between the ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made

Answer 1

a.) Since , By ohm's law,

V = I*R

So, I = V/R

here, V = 22.1 V

and R = 15.2 mohm

So, I = 22.1/(15.2*10^-3) = 1453.95 ohm = 1.45*10^3 Amp

b.) currnt density(J) = I/A

here, area(A) = pi*(r)^2

and r = 8.2/2 mm = 4.10*10^-3

So, J = 1453.95/(pi*4.10*4.10*10^-6)

J = 27531627.54 amp/m^2

J = 2.75*10^7 amp/m^2

c.) since, resistance(R) = rho*l/A

So, rho = R*A/l

here, l = 3.60 m

rho = (15.2*10^-3*pi*4.1*4.1*10^-6)/3.60

rho = 2.23*10^-7 ohm*m = resistivity of the material of which the wire is made.