Interesting Magnetic Field Induction Problem

[Klaus von Faust]

Homework Statement

Suppose we have an isolated, long, narrow straight wire with low electrical resistivity. A constant current $I$ is sent through this wire. We know that if an electron is sent on a path which is perpendicular to the wire, towards it, with an initial speed $v_0$, and the electron starts moving from a distance $x$, the maximal approach distance is $\frac x 2$. We will analyse the wire from two different systems of reference, one being a non-moving(with respect to the wire) system, and the other being a reference system moving with a speed of $v'$ parallel to the wire in the direction of the current flow. Find out the magnetic field induction of the wire in both of the reference systems. Find the Lorentz Force acting on the electron in both reference frames, and their vectorial difference.

Homework Equations

$\vec F_L=q[\vec v \vec B]$
$I=nevS$
$\oint \vec B \cdot d \vec{l} = \mu_0 I$

The Attempt at a Solution

Using the circulation theorem, it is easy to find out the magnetic field induction in the first case
$B=\frac {\mu_0I} {2\pi r}$
That is what I did for the second case
$I=nevS$
$I'=neS(v-v')=I-neSv'$ therefore $I'=I(1-\frac v v')$
All I have to do now is to find the velocity of the electrons in the wire. I don't know $n$ and $S$ and don't know what can I do next. I need the induction to solve the next part

 

[TSny]

Besides the negative charge of the conduction electrons, there is also the positive charge of the atomic ions. In the first frame of reference, the positive charge is at rest. But in the second frame, the positive charge is in motion and contributes to the overall current in this frame. Your calculation of $I'$ has not taken this into account.

The problem statement seems odd to me. The information about the electron that is projected toward the wire doesn't appear to me to have anything to do with the question about the magnetic induction in the two frames of reference. Have you quoted the question statement exactly? Are there any additional parts of the problem that you have not given?

Are you supposed to include relativistic effects when transforming between the frames of reference?

 

[Klaus von Faust]

Besides the negative charge of the conduction electrons, there is also the positive charge of the atomic ions. In the first frame of reference, the positive charge is at rest. But in the second frame, the positive charge is in motion and contributes to the overall current in this frame. Your calculation of $I'$ has not taken this into account.

The problem statement seems odd to me. The information about the electron that is projected toward the wire doesn't appear to me to have anything to do with the question about the magnetic induction in the two frames of reference. Have you quoted the question statement exactly? Are there any additional parts of the problem that you have not given?

Are you supposed to include relativistic effects when transforming between the frames of reference?

You are supposed to ignore the relativistic effect. The information about the electron is needed to solve the task about Lorentz Force difference. Using the fact that positive charge carriers move as well:
$I'=en_-(v-v')S-en_+v'S=enS(v-2v')$ because the positive charges move antiparallel to the current, and $n_+=n_-=n$
What can I do next? I still need to find $v$

 

[TSny]

Using the fact that positive charge carriers move as well:
$I'=en_-(v-v')S-en_+v'S=enS(v-2v')$ because the positive charges move antiparallel to the current, and $n_+=n_-=n$
What can I do next? I still need to find $v$

In your class, is the direction of current defined in terms of "electron current" or "conventional current"? http://web.engr.oregonstate.edu/~traylor/ece112/beamer_lectures/elect_flow_vs_conv_I.pdf

It is common these days to use "conventional current". So, if the wire is oriented along the y-axis and if the current is in the positive y-direction, the conduction electrons move in the negative y-direction. The problem states that the second reference frame moves relative to the wire in the direction of the current. So, the second frame moves in the positive y-direction relative to the first frame.

It's not easy keeping track of all of the signs. But I don't believe you are getting the correct expression for $I'$.

 

[Klaus von Faust]

In your class, is the direction of current defined in terms of "electron current" or "conventional current"? http://web.engr.oregonstate.edu/~traylor/ece112/beamer_lectures/elect_flow_vs_conv_I.pdf

It is common these days to use "conventional current". So, if the wire is oriented along the y-axis and if the current is in the positive y-direction, the conduction electrons move in the negative y-direction. The problem states that the second reference frame moves relative to the wire in the direction of the current. So, the second frame moves in the positive y-direction relative to the first frame.

It's not easy keeping track of all of the signs. But I don't believe you are getting the correct expression for $I'$.

Ok, I found my mistake. I solved the problem and found out that the magnetic field induction vector is the same no matter the reference system. Thank you