Question

A solenoid 5 cm in diameter and 30 cm in length has 4000 turns and carries a current of 5 A. Calculate the magnetic flux through the circular cross-sectional area of the solenoid. Since this is a very long solenoid, you may use the simplified magnetic field formula for the infinite solenoid

Answer 1

First you need to calculate the magnetic field inside the solenoid:
B=(permeability of free space)(Number of turns/Length)Current
B=(1.25664 x 10^-6)(4000/0.3)(5)
B= 0.0838

Now you need to calculate the cross-sectional area of the solenoid
Since the diameter is 5cm, the radius is 2.5cm = 0.025m
then you calculate the area = pi(0.025)^2=1.96 x 10^-3

now the formula for the magnetic flux is Magnetic field*Area
=(0.0838) (1.96 x 10^-3) = 1.6.4 x 10^-4

Answer 2

B =

Answer 3

Flux = u * N * I*A/ L

= 4pi*10^-7 * 4000 * 5 * pi*0.025^2 / 0.3

= 1.64 *10^-4 Wb

Answer 4

First you need to calculate the magnetic field inside the solenoid:
B=(permeability of free space)(Number of turns/Length)Current
B=(1.25664 x 10^-6)(4000/0.3)(5)
B= 0.0838

Now you need to calculate the cross-sectional area of the solenoid
Since the diameter is 5cm, the radius is 2.5cm = 0.025m
then you calculate the area = pi(0.025)^2=1.96 x 10^-3

now the formula for the magnetic flux is Magnetic field*Area
=(0.0838) (1.96 x 10^-3) = 1.64 x 10^-4 Wb