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consider a long, straight wire of cross-sectional area $A$ carrying a current $i$. Let there be $n$ free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the current with a speed $v = \frac{i}{nAe}$ and separated from the wire by a distance $r$. The magnetic field seen by the observer is very nearly

My Answer:

Zero. Because current is $neAv$ where $v$ is drift velocity of electrons. Relative velocity between him and electrons is zero. So, no flow of charge through any cross-section according to him. So no current. So no magnetic field.

Actual answer:

$\frac{\mu\ i}{2\cdot\pi\cdot r}$ where $\mu$ is the permeabilty of free space.

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closed as not a real question by Qmechanic, David Z Jan 1 '13 at 7:46

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

And what question are you asking us? If it is "what did I do wrong" think about what's in the wire apart from electrons. – leftaroundabout Feb 20 '12 at 0:13
up vote 3 down vote accepted

Unlike an electron beam, a wire carrying a current contains positive charges as well, and these charges move with respect to the moving observer.

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Opposite to current=along with electrons. – Manishearth Feb 20 '12 at 4:17
@Manishearth: I stand corrected. – akhmeteli Feb 20 '12 at 5:47
You may want to un-edit the edit then =P – Manishearth Feb 20 '12 at 5:49

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