Magnetic field at a point due to a line of charge

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
7 replies · 2K views
vysero
Messages
134
Reaction score
0

Homework Statement


A wire carrying a current I in the positive x direction is located along the x axis. The wire is of finite length and is located between x = -L and x = L. Find the magnetic field at a field point located a distance a away from the wire.

Homework Equations


Biot-Savart law
Ampere's law

The Attempt at a Solution



Problem.png

[/B]
I solved the problem above and its the same as the professors so its correct. However, I was following along with a video I found on lassevrien's channel on YouTube:

and he attained a different answer for what I believe is the same problem. If you don't want to watch the video I have summarized it on the picture (lower right); he is using Ampere's law.

So I figure if I evaluated my answer from -∞ too ∞ I would get what he got but I am getting an undefined answer. Are these two methods of doing the same problem? Am I just evaluating wrong?
 
Attachments
  • Problem.png
    Problem.png
    49.8 KB · Views: 470
Physics news on Phys.org
vysero said:
and its the same as the professors so its correct
That is in general a very dangerous assumption: to err is human and professors are human
vysero said:
I am getting an undefined answer
How so ?
 
BvU said:
How so ?

I believe B would be undefined here:

$$B =\frac{μI}{4πa} (\frac {∞} {\sqrt{a^2 +∞^2}}\frac {∞} {\sqrt{a^2 +∞^2}}),$$
 
Divide numerator and denominator by ##L## before letting ##L## go off to ##\infty##
 
  • Like
Likes   Reactions: vysero
BvU said:
Divide numerator and denominator by ##L## before letting ##L## go off to ##\infty##

I am not sure if this is what you meant but here goes:

$$\lim_{x \rightarrow +\infty} {\frac {x} {\sqrt{a^2 +x^2}}}$$
$$\lim_{x \rightarrow +\infty} {\frac {(\frac{1}{x})x} {\sqrt{\frac{1}{x^2}}\sqrt{a^2 +x^2}}}$$
$$\lim_{x \rightarrow +\infty} {\frac {1} {\sqrt{(\frac{1}{x^2})(a^2 +x^2)}}}$$
$$\lim_{x \rightarrow +\infty} {\frac {1} {\sqrt{\frac{a^2}{x^2}}+1}}=1$$
 
Brilliant ! Now add 1 and 1 :wink: .

Well, almost brilliant: not $$ \lim_{x \rightarrow +\infty} {\frac {1} {\sqrt{\frac{a^2}{x^2}}+1}}=1$$ but $$
\lim_{x \rightarrow +\infty} {\frac {1} {\sqrt{{\frac{a^2}{x^2}}+1}}}=1$$