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Self inductance formula question

  1. Aug 28, 2009 #1
  2. jcsd
  3. Aug 28, 2009 #2


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    [itex]A[/itex] is the cross sectional area of the solenoid/telephone cord, and the magnetic field inside the solenoid is [itex]B=\mu_0 NI/l[/itex] so the flux through one loop in the solenoid is [itex]BA=\mu_0 N IA/l[/itex] so the total flux is [itex]\Phi= \mu_0 N^2 I A/l[/itex] and the inductance is just total flux divided by current.
  4. Aug 28, 2009 #3
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