Ratio magnetic induction between 2 solenoids

[b_andries]

We have 2 solenoids : S1 and S2
The coils are wind up as close as they can get and both have length L.
They are both made from the same material and have the same length.
They are both connected to a DV source.

S2 has a wire with a double diameter
The ratio between B1 and B2 then is?

I thought the answer here was 1/2 but it was 1/4
My thoughts here were:
B1= (µ . I . n) / L
B2= (µ . 4I . (n/2) ) / L
(µ and L are constant and we can scrape them)


The current in S2 is 4 times bigger than S1 because according to pouillet the resistance is 4 times smaller.
( Solenoid 1 : R1 = (ρ) . L / (d/2)² * π
Solenoid 2 : R2 = (ρ) . L / (2d/2)² * π )

But then again S2 has to have half the number of the coils than S1 because the wire is double the diameter.

So B1/B2

B1= (µ . I . n) / L
B2= (µ . 4I . (n/2) ) / L ==> 1/2

What am I doing wrong here?

thank you!

 

[tiny-tim]

hi b_andries!

The coils are wind up as close as they can get and both have length L.
They are both made from the same material and have the same length.
…
But then again S2 has to have half the number of the coils than S1 because the wire is double the diameter.

the question is a little misleading

i think the important point is that the lengths of the two wires are the same, so the number of turns must be the same

(yes, i agree that if they are wound in a single layer, then there will be less room for the thicker wire, so there must be less turns, but i think the information about length takes precedence)

 

[b_andries]

ok thanks than I understand it :)