Radial current in a hollow metal cylinder

In summary, we discussed finding an expression for the electric field strength inside a hollow metal cylinder with a given current flow and dimensions. Through manipulating equations and considering infinitesimal segments, we were able to find the resistance of a thin cylindrical shell and use it to determine the net resistance of the entire cylinder. This allowed us to find the potential difference and solve for the current passing through the cylinder. Additionally, we noted that the segments are in parallel connection, making it easier to calculate the total resistance.
  • #1
Tabiri
16
0

Homework Statement



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.

Homework Equations



J = I/A

J = [itex]\sigma[/itex]E

The Attempt at a Solution



I tried setting the two equations for J equal to each other, so I/A = [itex]\sigma[/itex]E. Then to get E by itself, E=I/(A*[itex]\sigma[/itex]). I then substituted in the area of a cross section of the cylinder, where A=(pi)r^2. This gave me E=I/([itex]\sigma[/itex][itex]\pi[/itex]r^2). The problem is this equation doesn't include L.
 
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  • #2
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
Well, the resistance would be R=L/Aσ, and since A=[itex]\pi[/itex]r^2, then R=L/σ[itex]\pi[/itex]r^2.

To relate it to the whole cylinder, would you just multiply the whole thing by L? Because R=L/σ[itex]\pi[/itex]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[itex]\pi[/itex]r^2/L^2, or E=(I)σ[itex]\pi[/itex]r^2/L^3 if I was correct in multiplying R by L previously to get the resistance of the whole cylinder.
 
  • #4
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
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
Tabiri said:
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.

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
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
Yes, you are in the right direction now. Proceed to find the segment's resistance and hence the net resistance.
 
  • #9
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[itex]\pi[/itex], so R=1/2[itex]\pi[/itex]rσ.

Then substituting into the equation E=I/RL, E=I/2[itex]\pi[/itex]r(L)σ. Does that look right?
 
  • #10
Tabiri said:
So since A=r(dθ)L, and R=L/Aσ, then by substitution L cancels and R=1/r(dθ)σ for the segment.

How does L cancel? What is the length along which the current travels?
The area is right though
 
  • #11
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
Tabiri said:
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.

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
Would the net resistance just be R=(b-a)/Lr2[itex]\pi[/itex]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[itex]\pi[/itex].
 
  • #14
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
Well, the area of the shell would be the length * circumference, so A=2[itex]\pi[/itex]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
Tabiri said:
Would the net resistance just be R=(b-a)/Lr2[itex]\pi[/itex]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[itex]\pi[/itex].

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 [itex]2\pi[/itex]
 
  • #17
Sunil Simha said:
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 [itex]2\pi[/itex]

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[itex]\pi[/itex]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[itex]\pi[/itex]σ. What am I doing wrong, that I get two different expressions for E?
 
  • #18
Tabiri said:
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[itex]\pi[/itex]σ

How is ΔV=EL? Isn't L the length of the cylinder? That is not the length along which the current moves.
 
  • #19
Sunil Simha said:
How is ΔV=EL? Isn't L the length of the cylinder? That is not the length along which the current moves.

Ah, so it would be ΔV=I(b-a). Then (b-a) cancels and everything works out, and it's E=I/Lr2[itex]\pi[/itex]σ.
 
  • #20
Tabiri said:
Ah, so it would be ΔV=I(b-a). Then (b-a) cancels and everything works out, and it's E=I/Lr2[itex]\pi[/itex]σ.

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

Enjoy physics :smile:
 
  • #21
Thank you so much for the help!
 
  • #22
Tabiri said:
Well, the area of the shell would be the length * circumference, so A=2[itex]\pi[/itex]L.
##A=2\pi rL##
Since the current flows radially outward, then I think I would use the radial cross section.
No, you want the area perpendicular to the flow, so it's ##A=2\pi rL##. The length of resistance through which this flows is dr, so dR = dr/##2\pi rL \sigma##.
Then since I=ΔV/R, I=(ΔV)(dr)σ.
You want dV = I dR = I dr/##2\pi rL \sigma##. Whence E = dV/dR = I /##2\pi rL \sigma##.
 
  • #23
electric field in a cylindrical conductor

you think of these thin cylindrical shells as resistors in series, since the same current flows through each shell, for which you add the resistances.
then dR=ρdr/(2πrL) where resistivity ρ=1/σ

now dV=IdR
and E=-dV/dr = -IdR/dr = -Iρ/(2πrL)

now you have to decide what the electric field is on the outer surface of the cylinder and use it to determine an additive constant.
 
  • #24
jopkoam said:
now you have to decide what the electric field is on the outer surface of the cylinder and use it to determine an additive constant.
I can understand how one might need an additive constant for a potential, but not for a field, surely?
 
  • #25
yes my mistake. no constant needed.
 

1. What is radial current in a hollow metal cylinder?

Radial current in a hollow metal cylinder refers to the flow of electric current that moves in a circular or radial path along the surface of the cylinder. This type of current is commonly seen in cylindrical conductors, such as wires or pipes, and is caused by an electric potential difference between the inner and outer surfaces of the cylinder.

2. How is radial current different from axial current?

Radial current and axial current are two different types of electric current that can occur in cylindrical objects. While radial current flows along the surface of the cylinder, axial current flows in a straight line parallel to the axis of the cylinder. Axial current is typically seen in long, thin conductors, while radial current is more common in shorter, thicker conductors.

3. What factors affect the strength of radial current in a hollow metal cylinder?

The strength of radial current in a hollow metal cylinder is affected by several factors, including the electric potential difference between the inner and outer surfaces of the cylinder, the conductivity of the metal, and the thickness and length of the cylinder. Additionally, the presence of any external magnetic fields or other nearby conductors can also affect the strength of the radial current.

4. How is radial current measured in a hollow metal cylinder?

Radial current can be measured using a variety of techniques, including using an ammeter to measure the flow of current through the cylinder or using a voltmeter to measure the potential difference between the inner and outer surfaces of the cylinder. In some cases, the strength of the radial current can also be indirectly measured by measuring the magnetic field surrounding the cylinder.

5. What are the practical applications of radial current in a hollow metal cylinder?

Radial current in a hollow metal cylinder has several practical applications, including in electrical circuits, where it is used to transfer power or information between components. It is also commonly used in electromagnets, where the circular flow of current creates a strong magnetic field. Additionally, radial current can be used in induction heating processes, where the heat generated by the current is used to melt or fuse metal objects.

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