Physics Asked on March 16, 2021
I was studying current in a wire, now a doubt raise me up. If we have, for example, two wires of the same length connected one to the end of the other, but one wire has the area two times the area of the other. What will be conserved here? (Conserved in the sense that the magnitude of the quantity will remain the same) The current or the density current?
$I = int J da$
I was thinking about it, and since J is independent of A, i would predict that J is conserved and so the currents magnitude changes in the discontinuity.
In the same way $V = RI$ implies, with R = pl/A, that the current is proportional to area, so technically we have two argument in prol of |J| maintain the same.
But, mathematically, the antiderivative is always continuous, even if the function to be integrate is discontinuity in certain special cases (as that, it is discontinuous in one point!). So, being I continuous, we need that I in the point which connects the wires be the same. But so, J need to changes!
$E$ (as $V$), is not constant in the circuit. It is 2 times greater where the wire is thinner. $J$ is also 2 times bigger there. The expression $V = RI$ translates as $E = rho J$ where $rho$ is the resistivity.
Looking at the first expression ($V = RI$), $V$ and $R$ changes while $I$ is constant.
Looking at the second expression ($E = rho J$), $E$ and $J$ changes while $rho$ is constant.
The current $I$ is conserved, not $J$.
Correct answer by Claudio Saspinski on March 16, 2021
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