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?

  • $\begingroup$ 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


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

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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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