Magnetic field inside a solenoid

[fisher garry]

I have a problem with the derivation above I don't get how

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

 

[PeroK]

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 $dx = \frac{r d\theta}{\cos \theta}$

 

[fisher garry]

well it is not from a textbook it is a document I recieved. Unfortunately I am a bit lost from

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

starting from

 

[PeroK]

well it is not from a textbook it is a document I recieved. Unfortunately I am a bit lost from
View attachment 254268

You're lost because that is wrong. Try what I posted.

 

[fisher garry]

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?

 

[256bits]

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.

 

[PeroK]

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.

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

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

 

[sk_astroman]

@fisher garry: can you please let us know which book you referred for above relation in your picture posted?