Answer: Ø = 0.034Wb
Explanation:
A solenoid is idealized as a certain number of current loops of diameter d in series.
• The current loops create a magnetic field. This field permeates each loop, ie, there is magnetic flux through each loop.
Determine the number of turns from the length and given winding rate.
• Calculate the strength of the magnetic field produced by the solenoid, then the flux through one turn (winding).
Please find the attached file for the solution.
The magnetic flux through the circular cross-sectional area of the solenoid is 9.8×10⁻⁹ Wb.
Given information:
diameter of the solenoid d = 2.5 cm so radius r = 1.25cm
number of turns N = 4800
current I = 2A
length of the solenoid L = 30cm
Magnetic flux is the number of magnetic field lines passing through a given closed surface. It provides the measurement of the total magnetic field that passes through a given surface area. mathematically the magnetic flux is defined as the product of the magnetic field intensity and the area of the closed surface. The magnetic flux through the cross-sectional area of the solenoid is given by:
[tex]\phi=\mu_o nIA[/tex]
where n = N\L and A is the area
[tex]\phi=\frac{\mu_o NIA}{L}\\\\\phi=\frac{4\pi \times10^{-7}\times 4800\times \pi\times(1.25\times10^{-2})^2}{30\times10^{-2}}\\\\\phi=9.8\times10^{-6}Wb[/tex]
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