Transposition of formula(inductance)

[NUMB_NUTS]

Hi i hope you can help me?
My question is:

Given the relationships:

B/h = U, H = IN/l and L = N x dphi/dI {ie, L = N x change in phi/change in I}.

show that the inductance of the coil is:

L = Nsquared/S Where S = l/uA ??

.l is lower case L for length
.U is absolute permeability.
your help would be most welcome please many thanks

 

[big man]

Have you got any ideas on how you might find the equation for the inductance of a solenoid?

 

[kyle8921]

Hi, thank you for reaching out for help with transposition of the formula for inductance. I would be happy to assist you with this question.

First, let's review the given relationships:

B/h = U
H = IN/l
L = N x dphi/dI

We can see that the first two relationships involve magnetic field strength (B) and magnetic flux density (H), while the third relationship involves inductance (L). In order to solve for L, we need to eliminate B and H from the equations.

To do this, we can rearrange the first equation to solve for B, giving us B = U x h. We can then substitute this into the second equation, giving us H = IN/(U x h). Next, we can rearrange the third equation to solve for dphi/dI, giving us dphi/dI = L/N. We can then substitute this into the second equation, giving us H = IN/(U x h) = IN/(U x h) x (L/N). Simplifying this, we get H = IL/(U x hN).

Now, we can substitute this value for H into the first equation, giving us:

B/h = U x (IL)/(U x hN)

Simplifying this, we get B/h = IL/(hN)

Next, we can rearrange this to solve for L, giving us L = B x hN/I. Finally, we can substitute the given values for B and hN, giving us L = (N x dphi/dI) x (hN/I). Simplifying this, we get L = N x (dphi/dI) x (hN/I).

Now, we can rearrange this to get L = N x (hN/I) x (dphi/dI). Since we know that hN/I is equal to 1/uA, we can substitute this into the equation, giving us L = N x (1/uA) x (dphi/dI). Simplifying this, we get L = N x dphi/dI x (1/uA).

Finally, we can rearrange this to get L = N x (1/uA) x dphi/dI. Since we know that 1/uA is equal to l/U, we can substitute this into the equation, giving us L = N x (l/U) x dphi/dI. Simplifying