Magnetic field inside a solenoid

  • #1
1576606983811.png
1576607004731.png


1576607180280.png

I have a problem with the derivation above I dont get how
1576607218544.png

Can someone derive this and illustrate this visually for example by using Figure 2 or using another drawing?
 

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  • #2
PeroK
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It looks wrong to me. The book should have ##\theta## going clockwise with ##\theta = 0## as the vertical, and be integrating from ##-\frac{\pi}{2} + \theta_0## to ##\frac{\pi}{2} - \theta_0##.

It looks like the book has two wrongs making a right.

PS with the above I get ##dx = \frac{r d\theta}{\cos \theta}##
 
  • #3
well it is not from a textbook it is a document I recieved. Unfortunately I am a bit lost from
1576610265489.png

and the equations afterwards. Could you derive how to get to:
1576610310258.png

starting from
1576610504513.png
 
  • #5
1576643791061.png


I have tried to illustrate my problem in the drawing above. Since ##r d\theta## is normal to the radius r and approximately linear since it is a short part of the bowlength the angle between dx and ##r d\theta## should be the same as the angle between r and x that makes cosinus. But what if the fraction ##\frac{r d\theta}{dx}## and the ##\frac{r }{x}## is not the same? We don't know the length of ##r d\theta## measured up to r and the length of dx measured up to x?
 
  • #6
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View attachment 254289

I have tried to illustrate my problem in the drawing above. Since ##r d\theta## is normal to the radius r and approximately linear since it is a short part of the bowlength the angle between dx and ##r d\theta## should be the same as the angle between r and x that makes cosinus. But what if the fraction ##\frac{r d\theta}{dx}## and the ##\frac{r }{x}## is not the same? We don't know the length of ##r d\theta## measured up to r and the length of dx measured up to x?
Angles and infinitesimals - ugh.

Maybe this explanation helps.
##r d\theta## is the arc length, call that da,which for a radius r perpendicular to the axis would have dx = da.
A we deviate from the vertical, r increases in length, the arc length da is no longer parallel to the axis. We have to find the dx portion of da.
 
  • #7
PeroK
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View attachment 254289

I have tried to illustrate my problem in the drawing above. Since ##r d\theta## is normal to the radius r and approximately linear since it is a short part of the bowlength the angle between dx and ##r d\theta## should be the same as the angle between r and x that makes cosinus. But what if the fraction ##\frac{r d\theta}{dx}## and the ##\frac{r }{x}## is not the same? We don't know the length of ##r d\theta## measured up to r and the length of dx measured up to x?

As I said above, I'll take ##\theta## going clockwise from the vertical, so that ##\theta## and ##x## have the same sign.

##x = R\tan \theta, \ \ R = r\cos \theta, \ \ x = r\sin \theta##

##dx = R \sec^2 \theta d\theta = \frac{r d\theta}{\cos \theta}##
 
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