Question

Electric fields!

Let \(\vec{E}(\vec{r})=\int_a^b\dfrac{k\lambda}{\|\vec{r}-\vec{L}(t) \|^2}dt\), where \(\vec{L}\) is a piecewise continuous linear function with a constant linear charge density \(\lambda\) on it and k being the Columb constant. Is it possible to calculate the the electric field on \(\vec{r}\) which lies on \(\vec{L}(t)\)?

Answer 1

I am not sure whether this is possible or not because electric field in a conductor vanishes I think.

Answer 2

It is a simplification of a question that I am working on. The electric field is generated by moving the conductor in the magnetic field.

Answer 3

If the charge density is constant, then it would seem to me that there is no electric field in the conductor. I think I screwed this up.

Answer 4

Actually no, charge density is constant does not imply a lack of electric field.

Answer 5

inside conductor, there is no electric field, there is field outside the suraface.

Answer 6

I don't think this is the case if a straight rod is moving through a uniform magnetic field. There will be a potential difference between the two sides of the rod.

Answer 7

what case is it then? could you state the question more precisely? like what are the given conditions? I am thinking, a uniformly charged rod is just bend around and you are trying to calculate the electric field at certain point.

Answer 8

Okay. It was a high school examination question. A 0.050 m x 0.050 m square loop with discontinuity at the upper right hand corner is placed parallel to the plane of paper above a magnetic field of 25 mT pointing into the paper. The question ask for the emf induced in the square loop just before the whole loop enters the magnetic field.

Answer 9

I randomly scribbled something down the examination paper but I am not convinced that it can be solved without using any calculus.

Answer 10

In this case I think we should use Faraday's law

Answer 11

hmm .. if the loop is not closed then, there is no current

Answer 12

I think Faraday's law only applies to closed loop which the discontinuous square loop is clear not.

Answer 13

if you get enough thick materials, there will be eddy current, but no current in the loop.

Answer 14

The question didn't give the thickness of the loop, at least I didn't recall it did. So I will assume it is a infinitely thin wire.

Answer 15

*infinitesimally thin
ok no current

Answer 16

Technically I have to do the following to solve it. L(t) is the parametrisation of the open square loop.
\[
\vec{L}(t)=
\begin{cases}
(-t-0.050)i+0.050j&-0.099\leq t\leq 0.050\\
-tj & -0.050\leq t \leq 0\\
ti & 0\leq t\leq 0.050\\
0.050i + (t-0.050)j & 0.050\leq t\leq 0.100 \\
\end{cases}
\]
The magnetic field is defined as follows:
\[
\vec{B}(x,y,z)=
\begin{cases}
-25\times10^{-3}k&y<0\\
0&\text{otherwise}
\end{cases}
\]
The electric field has to satisfy the equation \(\vec{E}(\vec{r})=v\vec{B}(x,y,z)\). \(v=-0.990454\,\text{m}\,\text{s}^{-1}\) If I am not mistaken, the electric field has a formula of:
\[
\vec{E}(\vec{r})=\int_{-0.099}^{0.100}\dfrac{k\lambda(t)}{\|\vec{r}-\vec{L}(t) \|^2}dt
\]
for some linear charge distribution \(\lambda(t)\).
The potential difference (i.e. voltage) then can be found by:
\[
V=\int_L\vec{E}\cdot d \vec{l}
\]

How the f* is a high school student able to do that??????

Answer 17

Something is wrong with my post. The electric field only has to satisfy the equation
\[
\vec{E}(\vec{L}(t))=vB(x,y,z)
\]

Answer 18

But if \(\vec{r}=\vec{L}(t)\) for some t, then \(\vec{E}(\vec{r})\) cannot be evaluated right?

Answer 19

Is it true that if \[\lim_{\vec{r}\to\vec{L}(a)}\frac{\lambda(t)}{\|\vec{r}-\vec{L}(a)\|^2}=c\]
then the integral can be evaluated? But this implies \(\lambda(t)\)=0 at \(\vec{L}(t)=\vec{r}\) which is absurd. I must have done something wrong here.