A solenoid is 2.50 cm in diameter and 30.0 cm long. It has 300 turns and carries 12 A.
(a) Calculate the magnetic field inside the solenoid.
(b) Calculate the magnetic flux through the surface of a circle of radius 1.00 cm, which is
positioned perpendicular to and centered on the axis of the solenoid.
(c) The perimeter of this circle is made of conducting material. The current in the solenoid
uniformly goes from 12 A to 10 A in 0.001 seconds. What e. m.f. is generated in the
conducting material?
(d) Now, inside the solenoid we put a bar of steel, a ferromagnetic material. Steel has got a
magnetic permeability μm = 4000μ0 . If the current in the solenoid is 12 A, what’s the
total magnetic field in the steel bar?

a) B = 0.015 T

b) ФB = 4.71*10⁻6 W

c) emf = 7.89*10^-4 V

d) B = 60.31 T

Explanation:

(a) To find the magnitude of the magnetic field inside the solenoid you use the following formula:

     (1)

B: magnitude of the magnetic field

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 300

L: length = 2.50cm = 0.025 m

I: current = 12 A

You replace the values of the variables in the equation (1):

(b) The magnetic flux is given by:

A: area = π(0.01m)^2 = 3.1415*10^-4 m^2

(c) The induced emf is given by the following formula:

     (2)

Bf and Bi are the final and initial magnetic field. You use the equation (1) into the equation (2):

(d) With the ferromagnetic material inside the solenoid the magnetic field inside is modified as following:

where are the vectors ?