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mmiguel1
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I just had this problem on my Electromagnetics final. I want to know if I got this right and I can't find the problem with google.
Consider a conducting quarter-ring.
It can be envisioned as one piece of a hollow cylinder that has been cut into fourths down its length.
The original hollow cylinder would have had inner radius a, outer radius b, and length h.
In cylindrical coordinates the geometry of this quarter ring is: r \in [a,b]; \phi \in [0,\pi/2]; and z \in [0,h];
The material of interest has conductivity \sigma
What is the resistance from the \phi=0 end to the \phi=\pi/2 end?
For a conductor with linear geometry down it's length:
R = \frac{L}{\sigma A}
Where L = length and A = cross-sectional area;
G = \frac{1}{R}
I considered a slice of the conductor corresponding to a differential angle d \phi.
The slice is described by:
r \in [a,b]; \phi \in [\phi_0,\phi_0+d\phi]; and z \in [0,h];
Let the resistance of this slice be dR. Since the total conductor is a series combination of these slices, and since series resistances add,
R = \int dR
Within the slice of interest, dR, consider a sub-slice located at radius r with a < r < b.
The sub-slice is described by:
r \in [r_0,r_0+dr]; \phi \in [\phi_0,\phi_0+d\phi]; and z \in [0,h];
Let us call the resistance of this sub-slice dR'. Our original slice, dR, is a parallel combination of many dR' sub-slices. It is therefore more convenient to calculate the conductances for this part of the problem. Define dG = \frac{1}{dR} and dG' = \frac{1}{dR'}. Since parallel conductances add, we have
dG = \int dG'
As the differential values d\phi and dr tend to zero, the subslice of interest is linearized along its length.
We can therefore say
dG' = \frac{\sigma A}{L} = \frac{\sigma h dr}{r d \phi}
Where A = h dr and L = r d \phi
dG = \int_a^b \frac{\sigma h}{d \phi} \frac{1}{r} dr
dG = \frac{\sigma h}{d \phi} \ln \left (\frac{b}{a} \right )
dR = \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }
R = \int_0^{\frac{\pi}{2}} \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }
R = \frac{\pi}{2 \sigma h \ln \left ( \frac{b}{a} \right ) }
I was debating this question with my friends after the final. Each of us got different answers. I think my method is correct but I would like some verification.
Thanks!
Homework Statement
Consider a conducting quarter-ring.
It can be envisioned as one piece of a hollow cylinder that has been cut into fourths down its length.
The original hollow cylinder would have had inner radius a, outer radius b, and length h.
In cylindrical coordinates the geometry of this quarter ring is: r \in [a,b]; \phi \in [0,\pi/2]; and z \in [0,h];
The material of interest has conductivity \sigma
What is the resistance from the \phi=0 end to the \phi=\pi/2 end?
Homework Equations
For a conductor with linear geometry down it's length:
R = \frac{L}{\sigma A}
Where L = length and A = cross-sectional area;
G = \frac{1}{R}
The Attempt at a Solution
I considered a slice of the conductor corresponding to a differential angle d \phi.
The slice is described by:
r \in [a,b]; \phi \in [\phi_0,\phi_0+d\phi]; and z \in [0,h];
Let the resistance of this slice be dR. Since the total conductor is a series combination of these slices, and since series resistances add,
R = \int dR
Within the slice of interest, dR, consider a sub-slice located at radius r with a < r < b.
The sub-slice is described by:
r \in [r_0,r_0+dr]; \phi \in [\phi_0,\phi_0+d\phi]; and z \in [0,h];
Let us call the resistance of this sub-slice dR'. Our original slice, dR, is a parallel combination of many dR' sub-slices. It is therefore more convenient to calculate the conductances for this part of the problem. Define dG = \frac{1}{dR} and dG' = \frac{1}{dR'}. Since parallel conductances add, we have
dG = \int dG'
As the differential values d\phi and dr tend to zero, the subslice of interest is linearized along its length.
We can therefore say
dG' = \frac{\sigma A}{L} = \frac{\sigma h dr}{r d \phi}
Where A = h dr and L = r d \phi
dG = \int_a^b \frac{\sigma h}{d \phi} \frac{1}{r} dr
dG = \frac{\sigma h}{d \phi} \ln \left (\frac{b}{a} \right )
dR = \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }
R = \int_0^{\frac{\pi}{2}} \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }
R = \frac{\pi}{2 \sigma h \ln \left ( \frac{b}{a} \right ) }
I was debating this question with my friends after the final. Each of us got different answers. I think my method is correct but I would like some verification.
Thanks!
