# 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

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 = 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)

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