Respuesta :
Answer:
Magnetic flux, [tex]\phi=1.57\times 10^{-5}\ Wb[/tex]
Explanation:
It is given that,
Radius of the solenoid, r = 4 cm = 0.04 m
Length of the solenoid, L = 4 m
No of turns, N = 8000
Current, I = 5 A
Radius of the plane circular surface, r' = 2 cm = 0.02
Area of the circular surface,
[tex]A=\pi r'^2[/tex]
[tex]A=\pi (0.02)^2=0.00125\ m^2[/tex]
The magnetic flux through this surface is given by :
[tex]\phi=B\times A[/tex]
B is the magnetic field of the solenoid
[tex]\phi=\mu_o\dfrac{N}{L}I\times A[/tex]
[tex]\phi=4\pi \times 10^{-7}\times \dfrac{8000}{4}\times 5\times 0.00125[/tex]
[tex]\phi=1.57\times 10^{-5}\ Wb[/tex]
So, the magnetic flux through this surface is [tex]1.57\times 10^{-5}\ Wb[/tex]/. Hence, this is the required solution.
