A solenoid with 300 turns has a radius of 0.040 m
and is 40 cm long. If this solenoid carries a current of 12 A, what is the
magnitude of the magnetic field at the center of the solenoid? 
The magnitude of the magnetic field at the center [on the axis]of a solenoid is easily calculated with the equation B = μnI, n = N/L.
B = magnetic field
μ = a constant (permeability constant)
N = number of turns
L = length
I = current
The magnetic field at the center of a solenoid is directly proportional to the number of turns of the wire and to the current (as well as the relevant permeability constant). It is inversely proportional to the Length of the solenoid. This means a shorter solenoid with more turns has a greater magnetic field.
B = 4πx10^(-7) x (300/0.4) x 12 = 11 mT