Magnetic field of a long current-carrying coil

           B = μo  N/ℓ  I

  • B    magnetic induction in T
  • μo    magnetic permeability in vacuum (air)
  • μo = 4π x 10-7 Hm-1  
  • N     number of turns
  • ℓ       length coil in m
  • I       electric current in  A


         A coil contains 200 turns and has a length of 40 cm.

         The electric current is 0.30 A

         Find the magnetic induction in the coil

         B = μo ( N/ℓ) I

         B = ( 4π x 10-7)(200/0.40) ( 0.30) = 9.4 x 10-5 T

         We find the direction of the magnetic induction with the Right-Hand-Rule

  • fingers of the right hand in the direction of I
  • the thumb points in the direction of the magnetic  induction B ( see the figure above)


        IN the coil the field lines are directed from  Z to N

So at the left side of the coil is the magnetic N pole and at the right side the magnetic S pole              

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