I have a physics problem that states

|| What is the magnetic field at the position of the dot in Figure ex $32.5$ ? Give your answer as a vector.

Ex 32.5

The correct solution is (according to the book) using The Biot-Savart law, Solution

My question is why is the book choosing to use $\sin(135^\circ)$ ? I understand this has something to do with perspective and convention. My assumption is convention would say the angle of theta starts from the positive $x$ axis. I want to understand why this solution uses the angle of theta from what appears the negative $x$ axis. As $\sin(135^\circ)$ is noted from WolframAlpha below.

Sin(135 degrees) from WolframAlpha

  • $\begingroup$ You don't show where you determined $\,\theta\,$. For the $\,B=\,$ calculation it's enough to note that $\,\sin 135^\circ = \sin 45^\circ\,$. $\endgroup$ – dxiv Mar 11 '18 at 23:12
  • $\begingroup$ I am not sure what you are asking me to add. $\endgroup$ – user136952 Mar 11 '18 at 23:14
  • $\begingroup$ If you are asking why $\,\theta = 135^\circ\,$ rather than $\,45^\circ\,$ then that's impossible to answer since the question doesn't define $\,\theta\,$ and doesn't show how it was derived. $\endgroup$ – dxiv Mar 11 '18 at 23:15
  • $\begingroup$ Hmmm, this could be a physics thing. How is the formula derived and what do the terms mean? I don't know. It doesn't matter and it is perspective. But you are correct "First quadrant where $x > 0, y >0$ traditionally cooresponds to angles between $0$ and $90$ and and the angles for $80$ to $180$ are traditional in "second quadrant" where $x < 0; y > 0$. $\endgroup$ – fleablood Mar 11 '18 at 23:18
  • 1
    $\begingroup$ "Why did they choose to use sin(135) instead of sin(45) when sin(45) = sin(135)?" I really think only a physicist in context can answer that. What does cm in negative values mean. Is there a pragmatic reason to start counting at (-1, 0) rather than (1,0) and to go clockwise rather than counter clockwise? $\endgroup$ – fleablood Mar 11 '18 at 23:21

Because the field produced by a charge q moving in direction v at position r follows the law

B = muo/4pi * q/|r|^2 * v x r

where B, v, and r are all vector quantities and x is the cross-product. The magnitude of a cross-product AxB is equal to the magnitudes of A and B multiplied by the angle between vectors A and B.

So in the diagram, you show, you want to measure the angle theta relative to the direction of the velocity vector (which in this case is upwards). If you rotated the whole diagram by any arbitrary angle, you should still use the same theta value - it is independent of the x and y axes you use.

  • $\begingroup$ So relative to the diagram they are measuring sin(135) from the -y axis? $\endgroup$ – user136952 Mar 11 '18 at 23:31
  • $\begingroup$ No, it’s from the direction of the green arrow, which happens to point in the positive $y$-direction. $\endgroup$ – amd Mar 11 '18 at 23:35
  • $\begingroup$ I see so if rotate that direction of the vector relative to the x axis (counter clockwise), then I have sin(135), what about the counter example? Could I rotate it to the right relative to the x axis? $\endgroup$ – user136952 Mar 11 '18 at 23:40
  • $\begingroup$ You need to use MathJax to format your posts. $\endgroup$ – gen-z ready to perish Mar 11 '18 at 23:50
  • $\begingroup$ @user136952 if you curl the fingers of your right hand in the direction you 'rotate' to measure the angle between two vectors, your thumb points in the direction of the cross-product vector. From your diagram, rotating 135 degrees anti-clockwise results in thumb up (and +ve value for sin(135)). Measuring clockwise would be thumb down (and -ve for sin(225)). So in your diagram (with a +ve charged proton) field is either +283 attoT out of the page, or -283 attoT into the page (which are both the same thing). Always use your right hand and you will get the physics convention right. $\endgroup$ – Penguino Mar 12 '18 at 0:38

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