Respuesta :
Answer:
Magnetic flux, [tex]\phi= 3.56\times 10^{-4}\ Wb[/tex]
Explanation:
Given that,
Number of turns in the solenoid, N = 250
Current in the solenoid is, I = 9 A
Radius of the solenoid, r = 0.075 m
Length of the solenoid, l = 0.14 m
The magnetic flux is given by :
[tex]\phi=BA[/tex]
B is the magnetic field
A is the area of cross section
[tex]\phi=\mu_oI\dfrac{N}{l}\times \pi r^2\\\\\phi=4\pi \times 10^{-7}\times 9\times \dfrac{250}{0.14}\times \pi (0.075)^2\\\\\phi=3.56\times 10^{-4}\ Wb[/tex]
So, the magnetic flux through the circular cross-sectional area at the center of the solenoid is [tex]3.56\times 10^{-4}\ Wb[/tex]. Hence, this is the required solution.
Answer:
Explanation:
number of turns, N = 250
current, i = 9 A
radius of solenoid, r = 0.075 m
length of solenoid, l = 0.14 m
number of turnsp er unit length, n = N / l  = 250 / 0.14 = 1785.7
The magnetic field due t the current carrying solenoid
B = μoni
B = 4 x 3.14 x 10^-7 x 1785.7 x 9
B = 0.02 Tesla
The magnetic flux is given by
Φ = B x A
where, A is the area of crossection of the solenoid
A = 3.14 x r x r = 3.14 x 0.075 x 0.075 = 0.0177 m²
Ф = 0.02 x 0.0177
Ф = 3.57 x 10^-4 Weber
