# Radial current in a hollow metal cylinder

1. ### Tabiri

16
1. The problem statement, all variables and given/known data

A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity sigma. The current I is radially outward from the inner surface to the outer surface.

Find an expression for the electric field strength inside the metal as a function of the radius r from the cylinder's axis.

2. Relevant equations

J = I/A

J = $\sigma$E

3. The attempt at a solution

I tried setting the two equations for J equal to each other, so I/A = $\sigma$E. Then to get E by itself, E=I/(A*$\sigma$). I then substituted in the area of a cross section of the cylinder, where A=(pi)r^2. This gave me E=I/($\sigma$$\pi$r^2). The problem is this equation doesn't include L.

2. ### Sunil Simha

262
If you were to consider a thin radial section of the cylinder (so that the cross section would be almost uniform throughout), what would be its resistance?

Once you have figured that out, how would it be related to the resistance of the entire cylinder?

(Hint: the voltage across the ends of any such segment would be the same)

3. ### Tabiri

16
Well, the resistance would be R=L/Aσ, and since A=$\pi$r^2, then R=L/σ$\pi$r^2.

To relate it to the whole cylinder, would you just multiply the whole thing by L? Because R=L/σ$\pi$r^2 would just be the resistance for the cross section, so if you multiplied it by L then you'd get the resistance for the whole thing.

The only other thing I can think of is take the equations I=ΔV/R and E=ΔV/L. Manipulating equations, ΔV=I/R, substituting in for ΔV in the other equation you get E=I/RL, and substituting in for R you'd get E=(I)σ$\pi$r^2/L^2, or E=(I)σ$\pi$r^2/L^3 if I was correct in multiplying R by L previously to get the resistance of the whole cylinder.

4. ### Sunil Simha

262
No no, I think you misunderstood me. Check out my attachment. I have marked out a segment right? Imagine that segment to be subtending an infinitesimal angle at the center. What would its resistance from the inside to outside be?

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5. ### Tabiri

16
Well, if it's an infinitesimally small angle, then wouldn't the cross section just be a rectangle, and the area would just be outer radius - inner radius * L, so (b-a)L.

Edit with the resistance: R=L/(b-a)Lσ, and L would cancel, so it would just be R=1/(b-a)σ.

6. ### Sunil Simha

262
The current is flowing normal to the lateral surface. So what then will be the cross section area?

(assume that the angle subtended is dθ)

7. ### Tabiri

16
Would it have to do with arc length? Assuming θ is in radians, then the arc length would be s=r(dθ). Then the cross section area that the current passes through would be A=r(dθ)L.

8. ### Sunil Simha

262
Yes, you are in the right direction now. Proceed to find the segment's resistance and hence the net resistance.

9. ### Tabiri

16
So since A=r(dθ)L, and R=L/Aσ, then by substitution L cancels and R=1/r(dθ)σ for the segment. If we were to extend that all the way around the cylinder, then the total θ would be 2$\pi$, so R=1/2$\pi$rσ.

Then substituting into the equation E=I/RL, E=I/2$\pi$r(L)σ. Does that look right?

10. ### Sunil Simha

262
How does L cancel? What is the length along which the current travels?
The area is right though

11. ### Tabiri

16
I was putting what I got for A, r(dθ)L into R=L/Aσ, which goes to R=L/Lr(dθ)σ. Unless, for resistivity, would it be R=(r-b)/Aσ? Because since the current radiates outwards from the inner surface to the outer surface, then the current would only exist in the cylinder. Then it would be R=(r-b)/Lr(dθ)σ.

12. ### Sunil Simha

262
Yes. That is the correct answer and reasoning. (only, instead of r-b it should be b-a)

Now what would be the net resistance of the cylinder?

(Hint: The segments are subjected to the same potential difference)

13. ### Tabiri

16
Would the net resistance just be R=(b-a)/Lr2$\pi$σ? Since dθ is just a small segment, then if it were to be rotated all the way around the cylinder, then the total θ would be 2$\pi$.

14. ### haruspex

12,483
You can make this easier by looking at a thin cylindrical shell, radius r, thickness dr.
What is its area? What is its length in the direction of current flow? What is its resistance therefore? How much current is passing through it? What does that tell you about the potential difference from r to r+dr?

15. ### Tabiri

16
Well, the area of the shell would be the length * circumference, so A=2$\pi$L. If it was a radial cross section, then the area of that would be dr*L. Since the current flows radially outward, then I think I would use the radial cross section. This would make the resistance R=L/Aσ = L/L(dr)σ = 1/(dr)σ. Then since I=ΔV/R, I=(ΔV)(dr)σ. I'm not really sure what that says about the potential difference, though.

16. ### Sunil Simha

262
Yes that is correct. Now That you know the voltage, the current can be easily found.

(by the way, observe that all those segments are in parallel connection and hence you can simply sum up the angle to $2\pi$

17. ### Tabiri

16
Well, I=ΔV/R, so ΔV=RI. Also, E=ΔV/L, so ΔV = EL. Then RI=EL, and E=RI/L = I(b-a)/(L^2)r2$\pi$σ

Since electric field was what I was trying to find originally. The strange thing is what happens if I do it a different way. J = I/A = σE, so E=I/Aσ. Then it would be E=I/Lr2$\pi$σ. What am I doing wrong, that I get two different expressions for E?

18. ### Sunil Simha

262
How is ΔV=EL? Isn't L the length of the cylinder? That is not the length along which the current moves.

19. ### Tabiri

16
Ah, so it would be ΔV=I(b-a). Then (b-a) cancels and everything works out, and it's E=I/Lr2$\pi$σ.

20. ### Sunil Simha

262
ΔV=E(b-a). I guess, that was unintentional on your part.

Enjoy physics