Is There a Displacement Current in a Wire with Increasing Current?

AI Thread Summary
The discussion centers on whether a displacement current exists in a wire with increasing current, specifically focusing on the relationship between conductivity and displacement current. Participants analyze the equation Id = e0/σ ⋅ dI/dt to demonstrate this relationship, while addressing the challenge of eliminating variables like length and area from their calculations. There is clarification on the units of conductivity (σ) and resistance (Ω), emphasizing that conductivity is measured in mhos/m and resistivity is the inverse. The conversation highlights the importance of understanding these units to solve the problem effectively. Ultimately, the participants work towards a clearer understanding of the displacement current concept in the context of increasing current in a wire.
Jamie_Pi
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Homework Statement


A wire with conductivity σ carries current I. The current is increasing at the rate dI/dt. Show that there is a displacement current in the wire equal to e0/σ⋅dI/dt

Homework Equations


Id = e0⋅dφ/dt
dφ/dt = dE/dt ⋅ A (This is usually true, I'm not sure if it's useful in this case)
dV/dt=dI/dt⋅R (A changed version of Ohm's law)
R=1/σ
E=V/l

The Attempt at a Solution


I started by saying that dV/dt = dI/dt⋅1/σ, in which case the equation might look like:
Id = e0/(σ⋅l)⋅dI/dt⋅A

This looks like I'm going in the right direction, but I'm not sure how to get rid of l (because I don't know the length of the wire, and presumably it doesn't matter) or A (I think that the area of a wire is supposed to be nominal). Any tips that you have would be super great!
 
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I'm trying a new formula, but I've come to a point of stagnation:

dφ/dt = dI/dt ⋅ 1/σ

φ has units vm
so dφ/dt must have units vm/s
1/σ is the same thing as Ω, which has units v/A
and dI/dt has units A/s
vm/s = A/s⋅ v/A

the only problem I have now is that unit of length, which I don't know how to get rid of. I'm not sure what it represents. Any ideas?
 
Jamie_Pi said:
1/σ is the same thing as Ω, which has units v/A

Doesn't σ have units of mhos/m while Ω has units of mhos-1 or reciprocally σ ( 1/ohms⋅m ) while Ω (ohms)?
 
gleem said:
Doesn't σ have units of mhos/m while Ω has units of mhos-1 or reciprocally σ ( 1/ohms⋅m ) while Ω (ohms)?

Oh, I think that you're right. That's good, that just clears up the whole thing then.
 
gleem said:
Doesn't σ have units of mhos/m ... or reciprocally σ ( 1/ohms⋅m )
You mean mho.m and m/ohm, right?
 
The unit of conductivity is mhos/m. A mho is 1/ohms so conductivity is 1/ohm⋅m.. Resistivity ρ =1/σ and has units of ohm⋅m or m/mho.
 
gleem said:
The unit of conductivity is mhos/m. A mho is 1/ohms so conductivity is 1/ohm⋅m.. Resistivity ρ =1/σ and has units of ohm⋅m or m/mho.
Sorry, my mistake. I was thinking of resistance of a wire as being so many ohms per metre, but that's not what resistivity means; it's a bulk property..
 
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