# Electric field in a cylindrical conductor

1. Nov 24, 2014

### Nikitin

1. The problem statement, all variables and given/known data
Problem 1c from here: http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Exam_tfy4240_Dec_2013.pdf

2. Relevant equations
Maxwell's equations

3. The attempt at a solution
According to the solutions, the electric field is ZERO everywhere because it's "magnetostatics" and because the net charge is zero everywhere. http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Solution_tfy4240_Dec_2013.pdf

But, in that case, what about the electric field (given by Ohm's law) which is driving the current in the first place? That one is certainly not zero. Or does it "converge to zero" since the conductor is defined to be "very long" and ohm's law gives $\lim_{|d| \to \infty} IR = \vec{E} \cdot \vec{d}$

Last edited: Nov 24, 2014
2. Nov 24, 2014

### Vagn

Isn't the question asking for the radial electric field (i.e. the E-field in the x-y plane), the field driving the current would be along the wire not radially.

3. Nov 24, 2014

### Nikitin

No it's asking for the Electric field $\vec{E}(\vec{r})=\vec{E}(\vec{r_{||}})$, i.e. E as a function of radial r.

4. Nov 24, 2014

### Vagn

Part b asks for the magnetic field as a function of distance from the z-axis, where the current flows parallel to z. It then asks you to repeat the process for the electric field, which would imply it is the electric field perpendicular to the current.
In the solutions to part b, the magnetic field is given as a function of the radial component only, the question is similarly explained in the footnote at the bottom of the question page, where it states 'Obtain the electric field E(r) as a function of r||=|r|||' for the same regions.

5. Nov 24, 2014

### Nikitin

they asked for the magnetic field as a function of radial distance. i don't think they said anything about finding the magnetic field-component perpendicular to the current.

Regardless, wouldn't the electrical field inside a conducting wire like this approach zero anyway in the limit its length going towards infinity? I mean, electrical field times length equals difference in potential, and if voltage remains constant while length diverges, then the electrical field must approach zero.

6. Nov 24, 2014

### vela

Staff Emeritus
As the wire gets longer, the resistance R would increase, so to keep the current fixed at I, you'd have to increase the potential difference V. In the microscopic view, you'd have $\vec{J} = \sigma\vec{E}$. Neither the current density nor the conductivity depend on the length of the wire, so $\vec{E}$ doesn't depend on the length either.

You're right that in a real wire, there'd be an electric field directed along the wire to keep the charges moving. You're supposed to use the approximation that this is an ideal conductor, so there's no resistance. Once the current is established, no electric field is needed to keep it moving.