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The question is deceptively simple.

Suppose I have a uniform circular wire in which I have created the a E-field by some mean. We that without the source the electric field should not be there, so if we calculate the time dependence of the current then it should decay with time.

Suppose at time t=0 s, the current in a closed wire is $I_o$. Assuming there was a electric field which caused it then

$$ E = \frac{I\rho}{A} $$

The energy stored in the E-field is, $U_o = \tau P$ where $\tau=\frac{1}{2} \epsilon_o \rho$ and $P=I^2 R$. (Since $U_o = \frac{1}{2} \epsilon_o E^2 \times (LA)$ where L-> length and A-> Area of wire.)

Now we know the wire is losing energy by Joule heating, i.e. $\frac{dU_o}{dt}=-P$ ==> $$\frac{dP}{dt}=-\frac{P}{\tau}$$. Solving this gives,

$$ P=P_o e^{-t/\tau} $$

or

$$ I=I_o e^{-t/{2\tau}} $$

So the current decreases exponentially. Good! It is all fine and dandy until I think of case of superconductors where $\rho=0 \rightarrow \tau=0 \rightarrow I = 0$ for any time $t\neq0$.

But then one can also question whether the equation of energy density in Electric field be applied Since the dielectric constant is equal to $\infty$ for conductors (metals).

So my question is where I am wrong or more appropriately what wrong steps I have made in my calculation.

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  • $\begingroup$ If $R = 0$ then the fact that $P = 0$ tells you nothing about $I$ $\endgroup$ – By Symmetry Aug 21 '15 at 1:10
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So my question is where I am wrong or more appropriately what wrong steps I have made in my calculation.

When you lower mathematically the resistance while fixing the current, this means electric field decreases, but magnetic does not.

In the limit $\rho\rightarrow 0$ all the initial energy is in the magnetic field and can be expressed through self-inductance $L$ and current $I$ as $\frac{1}{2}LI^2$. If you repeat your calculation of $I(t)$, you will find that the lower the resistance, the longer it takes for the current to decay.

But then one can also question whether the equation of energy density in Electric field be applied Since the dielectric constant is equal to ∞ for conductors (metals).

The original formula for electric field energy with vacuum permittivity is the correct one, do not put $\epsilon$ of medium. The latter epsilon makes sense in the electric field energy only for non-dissipative linear medium, where the polarization term can be taken into definition of field energy. It is a good approximation to use when visible light passes through water or glass, but it is not possible when current decays in metal.

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  • $\begingroup$ You were right my mistake was to ignore the magnetic field energy (which I deliberately did). By taking into account the magnetic field energy the current, surprisingly comes out in the form of the exponential form which at $\rho -> 0$ becomes a constant current. $\endgroup$ – The Imp Aug 21 '15 at 7:40
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The problem is in your assumptions. If you have an E-field in the loop then the potential (voltage) would be increasing as you go around it.

If you have a current in the wire, there must be a magnetic flux through the loop. The magnetic field has energy, which you are not accounting for.

The magnetic field cannot penetrate a superconductor, so it is "trapped" in the loop and the current is constant.

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  • $\begingroup$ I know that as a matter of fact I don't even want to restrain my E-field to be conservative in nature for the sake of the question. $\endgroup$ – The Imp Aug 20 '15 at 17:18

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