The formula for the magnetic field of a solenoid is given by,
B = μoIN / L
Where,
N = number of turns in the solenoid
I = current in the coil
L = length of the coil.
Please note that the magnetic field in the coil is proportional to the applied current and number of turns per unit length.
Solved Examples
Example 1
Determine the magnetic field produced by the solenoid of length 80 cm under the number of turns of the coil is 360 and the current passing through is 15 A.
Solution:
Given:
Number of turns N = 360
Current I = 15 A
Permeability μo = 1.26 × 10−6 T/m
Length L = 0.8 m
The magnetic field in a solenoid formula is given by,
B = μoIN / L
B = (1.26×10−6 × 15 × 360) / 0.8
B = 8.505 × 10−3 N/Amps m
The magnetic field generated by the solenoid is 8.505 × 10−4 N/Amps m.
Example 2
A solenoid of diameter 40 cm has a magnetic field of 2.9 × 10−5 N/Amps m. If it has 300 turns, determine the current flowing through it.
Solution:
Given:
No of turns N = 300
Length L = 0.4 m
Magnetic field B = 2.9 × 10−5 N/Amps m
The magnetic field formula is given by
B = μoIN /L
The current flowing through the coil is expressed by
I = BL / μoN
I = (2.9×10−5 x 0.4 )/ (1.26×10−6 × 300)
I = 30.6 mA