Question

A 10.0 cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50.0 turns is wound around the solenoid at its center.

a) What is the mutual inductance of the combination of the two coils?

Answer 1

the mutual inductance  M = μ₀ N1*N2*A/L

 the No. of turns in first solenoid  N₁    =   800

 the No. of turns in second solenid   N₂   =   50.0

  the Length of solenoid   L   =   10.0cm =  0.10   m

  the Area of cross section   A   =   π* r²

         =   3.14* ( 0.400 * 10^-2 / 2)²

          =   1.256* 10^-5   m²

   M   =   4π* 10^-7 * 800 * 50.0 * 1.256 * 10^-5 /0.10

         =   6.31* 10^-6   H
   M = 6.31   ${\rm {\mu H}}$