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Thanks for the reply. What is the magnetic path l?It looks like it's just [itex]L = \mu \mu_0 N^2 A/l[/itex] where [itex]\mu_0 \simeq 1.26 E-6[/itex] is folded in with the conversion from meters to inches to give the combined constant of 3.2E-8.
No, the geometric factor A/l just has units of length. In SI units you use m^2 and m to give A/l in meters. Then of course you use [itex]\mu_0[/itex] in H/m (1.26E-6 H/m).Thanks for the reply. What is the magnetic path l?
[tex]L_0=\frac {3.2A\mu N^2}{10^8\times l}[/tex]
Also where is 3.2EE2 come from. [itex]\mu_0\;[/itex]=1.256EE-6 won't get 3.2EE2.
Thanks
Alsn
Yeah, it's the mean flux path length.Thanks, so it's just going from meter to inches.
So what is the path length [itesx]l[/itex]? It was given 4.5 inches, where is this come from? Is this the length of the coil?
No, the equation you post for the solenoid is not for the total inductance, it is for the inductance per unit length. It should read :With that [itex]L_0\;[/itex] should be:
[tex] L_0= \frac {3.2A \mu_0\mu}{10^8}\frac{N^2}{l^2}\;\hbox{ instead of }\;L_0= \frac {3.2A \mu_0\mu}{10^8}\frac{N^2}{l}[/tex]
Thanks, do you go to sleep? I put in my sleeping in between these posts and you're still here!!!No, the equation you post for the solenoid is not for the total inductance, it is for the inductance per unit length. It should read :
[tex]\frac{L}{l}=\mu_0\mu n^2 A [/tex]