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Problem 58 Hard Difficulty

A straight, cylindrical wire lying along the $x$ axis has a length $L$ and a diameter $d .$ It is made of a material described by Ohm's law with a resistivity $\rho$ . Assume potential $V$ is maintained at the left end of the wire at $x=0$ . Also assume the potential is zero at $x=L .$ In terms of $L, d$ , $V, \rho,$ and physical constants, derive expressions for (a) the magnitude and direction of the electric field in the wire, (b) the resistance of the wire, (c) the magnitude and direction of the electric current in the wire, and (d) the current density in the wire. (e) Show that $E=\rho J$.

Answer

a. $\vec{E}=\frac{V}{L} \hat{i}$
b. $R=\frac{4 \rho L}{\pi d^{2}}$
c. $I=\frac{\pi d^{2} V}{4 \rho L}$
d. $\frac{V}{\rho L} \hat{i}$
e. $\rho \vec{J}=\vec{E}$

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Video Transcript

So for part A here, we know that voltage is equal to the electric field times the distance, which in this case is L. The length of the wire. And so electric field is going to be equal to voltage over length. And that's going to be to the right so that I had direction for party. We have the equation. Resistance is equal to the resisted ity times the length over the area, so pi times diameter over two squared that could be simplified to give four times resisted many times the length over pi times diameter squared then for part C, we have current is equal to voltage over resistance. So plugging in what we got for a part B, we find that that equal to voltage times the denominator of the resistance. So pi diameter squared over the numerator of the resistance. So for time is the recessive ity times the length. Then for part D, we have the current density, which is equal to the current over the area. Yes, that's going to what we got from part c. The times pi d squared over four times Reason city times length over the area. So pi times diameter over two squared and simplifying this, we find that the current density is equal to voltage over three sensitivity times the length. And then finally, for Part E, we can simplify out the resistive ity times the current density using what we found in Part D as three sensitivity time is the voltage over the reason City times the length the's resisted ease. Cancel out. That gives us just voltage over length, but that's equal to the electric field, as we already used in part a.

St. Olaf College
Top Physics 102 Electricity and Magnetism Educators
Elyse G.

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Marshall S.

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