Question #108179
A straight, cylindrical wire lying along the x axis has a length of 0.500 m and a diameter of 0.200 mm. it is made of a meterial described by Ohm's law with a with a resistivity of p=4.00 *10^-8 Ohm . m. Assume a potential of 4.00 V is maintained at the left end of the wire at x=0. Also assume V=0 at x=0.500 m. Find a) the magnitude and direction of the electric field in the wire , b) the resistance of the wire, c) the magnitude direction of the electric current in the wire, and d) the current density in the wire. e) Show that E=pJ
1
Expert's answer
2020-04-08T10:55:44-0400

a)


E=400.50=8.00VmE=\frac{4-0}{0.5-0}=8.00\frac{V}{m}

Direction: rightwards (positive x-direction).

b)

R=4.001080.50.25π(0.0002)2=0.637ΩR=4.00\cdot10^{-8} \frac{0.5}{0.25\pi(0.0002)^2}=0.637\Omega

c) Direction: rightwards (positive x-direction).


I=VR=40.637=6.28 AI=\frac{V}{R}=\frac{4}{0.637}=6.28\ A

d)


J=6.280.25π(0.0002)2=2.00108Am2J=\frac{6.28}{0.25\pi(0.0002)^2}=2.00\cdot 10^8\frac{A}{m^2}

e)


E=ρJ=(4.00108)(2.00108)=8.00VmE=\rho J=(4.00\cdot10^{-8})(2.00\cdot 10^8)=8.00\frac{V}{m}


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