Magnetic field of a long current-carrying coil

           B = μ_o  N/ℓ  I

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

        Example

         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 = μ_o ( 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)

        Remark:

        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