Derivation of magnetic field of a Solenoid: Biot savart law

  • #1

Main Question or Discussion Point

Hello,

I have seen that biot savart's law works for infinitely narrow wires:
"The formulations given above work well when the current can be approximated as running through an infinitely-narrow wire."

When I wanted to derive the magnetic field of a solenoid, I had to do this substitution:
##n_o = N/L##

## k = n_o dx ##
Where k is the number of turns per dx.. But shouldn't K be an integer? so I can substitute it in the formula for circular coils. That means I have infinite number of turns and turn density of something like ## \frac{a}{dx} ## where a is an integer.

Is there is something wrong or that this is the idealization that we do to the solenoid? Wouldn't it be way off the correct value?

If you want the proof, http://nptel.ac.in/courses/122101002/downloads/lec-15.pdf
Page 8, Example 9.

Thank you in advance.
 

Answers and Replies

  • #2
BvU
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But shouldn't K be an integer?
No reason. If you have an 11 cm coil with 10 turns, you have 90.91 turns/m.

Note: never, never ever write something like "##
\frac{a}{dx}## with ##a## finite".
##dx## is universally seen as an infinitesimal (something that goes to zero). So ##
\frac{a}{dx}## does not exist.
 
  • #3
No reason. If you have an 11 cm coil with 10 turns, you have 90.91 turns/m.
Don't we idealize a solenoid as a number of circular coils?

https://i.imgur.com/RCO3qcQ.png

If we take a dx piece of this solenoid and treat at as k number of coils,

The equation for a single coil is:
## B = \frac{ u_o i R^2}{2 ( R^2 + x^2)^{\frac{3}{2}}} ##
If dx has 2 turns then we multiply by 2, If it has k turns then we multiply by k. But a dx piece cant have a 2.5 or a fraction of a coil (It doesn't make sense), Can it?
That is why I said ## k = n_o dx ## has to be an integer.
 
  • #4
BvU
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Don't we idealize a solenoid as a number of circular coils?
We calculate the B field as if it were caused by a cylindrical sheet of current. That is a very good approximation (example 8 already indicates that).

If dx has 2 turns then we multiply by 2, If it has k turns then we multiply by k. But a dx piece can't have a 2.5 or a fraction of a coil
Again: do NOT use infinitesimals this way -- it will get you into trouble. A infinitesimal ##dz## piece has zero turns and nevertheless contributes to B with an infinitesimal contribution ##dB## proportional to ##{N\over L}##. It is only when you integrate ##dB## over a finite range in ##z## that you get a finite result. The derivation (it's not a proof) in the pdf is just fine.
 

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