Originally Posted by **AKM17** What happens to the magnetic field inside this solenoid, if you double the number of revolutions, simultaneously stretching the solenoid twice as much as the original length? Suppose that the applied voltage was kept constant, and the additional wire length for additional turns of the coil did not noticeably increase the resistance of the solenoid?
Who can give a detailed answer and explain why this works this way? |

If you just want a general description of Ampere's law applied to a solenoid, you can try these sites:

There's hyperphysics:

HyperPhysics
Wikipedia:

https://en.wikipedia.org/wiki/Solenoid
There's probably a lot more websites too, but I usually use those two. They explain the formula for the magnetic flux density of the solenoid as a function of the number of turns, N, current, i, and length, l.

$\displaystyle B = \mu_0 \frac{N i}{l}$

where $\displaystyle \mu_0$ is the permeability of free space. so comparing two solenoids with the following turns and lengths:

1: $\displaystyle N = N_1, l = L_1$

2: $\displaystyle N = N_2 = 2 N_1, l = L_2 = 2 L_1$

then

$\displaystyle B_1 = \mu_0 \frac{N_1 i}{L_1}$

$\displaystyle B_2 = \mu_0 \frac{N_2 i}{L_2} = \mu_0 \frac{\cancel{2} N_1 i}{\cancel{2} L_1}=

\mu_0 \frac{N_1 i}{L_1} = B_1$

so the magnetic flux density is the same.