1
$\begingroup$

in the formula

$$dB = \frac{\mu_0l ~|dl \times r|}{4 \pi r^3} $$

and the image
enter image description here

where dl is in y-z plane and dB is in x-y plane. the ring conductor is in y-z plane carrying current I in direction as mentioned
EDIT: also point p can move in the circular ring

EDIT 2:To clear the confusion...The dl vector is having (L alphabet) and current is I (i alphabet).

I want to know that is the angle between dl and r is 'Theta' ? how?

$\endgroup$
4
  • $\begingroup$ are you claiming that $\theta$ is the angle between $dI$ and $r$? $\endgroup$ Commented Mar 1, 2014 at 16:10
  • $\begingroup$ No, but I want to confirm that...because it is given in my text book and i cannot understand how ? $\endgroup$ Commented Mar 1, 2014 at 16:14
  • $\begingroup$ is $dI$ parallel to z-axis and perpendicular to y-axis? $\endgroup$ Commented Mar 1, 2014 at 16:18
  • $\begingroup$ No, not always because point p is arbitrary point on that circular conductor $\endgroup$ Commented Mar 1, 2014 at 16:19

3 Answers 3

1
$\begingroup$

Angle between $dl$ and $r$ at any point on the circular loop is $90^0$. Look at the below figure. enter image description here

$\endgroup$
1
  • 2
    $\begingroup$ Which software is used to make such figures? $\endgroup$
    – user5402
    Commented Mar 1, 2014 at 20:21
1
$\begingroup$

the angle between $dl$ and $r$ is $\pi/2$, which is not $\theta$.

$\theta$ is the angle between $r$ and y-z plane. if you know what is $x$ and $r$, then $\sin\theta=x/r$, where $x$ is the distance from the origin to the point of intersection of $r$ and x-axis, and $r$ is the distance from p point to the same place

UPDATE. $\theta$ is important because your $dB$ is at this angle to x-axis. so when you add up all $dB$ resulting from all points p, only the $\cos\theta dB$ parts will contribute, because the part of $dB$ which is in y-z plan will cancel each other. for each point p, there's is an opposite point on the ring, they cancel each other's $dB$ on y-z plane, but not on x-axis. that's why resulting $B$ will be on x-axis.

$\endgroup$
5
  • $\begingroup$ So, according to you x/r is constant ? let me update the image $\endgroup$ Commented Mar 1, 2014 at 16:29
  • $\begingroup$ what do you mean by constant? it is constant regardless which point p you choose. it is not constant with regards to which point on x-axis you're considering $\endgroup$ Commented Mar 1, 2014 at 16:30
  • $\begingroup$ Fix this answer because the angle is (pie)/2 and not (pie)/4 $\endgroup$ Commented Mar 1, 2014 at 17:00
  • $\begingroup$ @MukulKumar, good catch $\endgroup$ Commented Mar 1, 2014 at 17:01
  • 1
    $\begingroup$ $\pi$ is 180, so 90 is $\pi/2$ $\endgroup$ Commented Mar 1, 2014 at 17:48
1
$\begingroup$

The angle between $\vec {dl}$ and $\vec r$ is $2n\pi \pm\dfrac{\pi}{2}$ because the angle between them is the angle between the x-r plane and y-z plane.

$\endgroup$
6
  • $\begingroup$ Please reconsider the question $\endgroup$ Commented Mar 1, 2014 at 16:12
  • $\begingroup$ The vector r only stays in x-y plain in two cases in rest cases the vector shifts away(in four octants of graph where x is positive) hence the angle is not what you are saying $\endgroup$ Commented Mar 1, 2014 at 16:23
  • $\begingroup$ The point p is not on y-axis but is on the circle (arbitrary).That is why the angle is not (never) a constant but a variable $\endgroup$ Commented Mar 1, 2014 at 16:41
  • 1
    $\begingroup$ I understood now.Looks like I got a problem of 3D imagination. $\endgroup$ Commented Mar 1, 2014 at 16:47
  • 1
    $\begingroup$ @MukulKumar Ask for a mathematical prove in your question. you will get a nice answer. Btw choose the most appropriate anwser as accepted cya. $\endgroup$
    – user31782
    Commented Mar 1, 2014 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.