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mariarad [96]
2 years ago
11

A solenoid 4.0 cm in radius and 4.0 m in length has 8000 uniformly spaced turns and carries a current of 5.0 A. Consider a plane

circular surface (radius = 2.0 cm) located at the center of the solenoid with its axis coincident with the axis of the solenoid. What is the magnetic flux through this surface? (1 Wb = 1 T ⋅ m2)Group of answer choices
Physics
1 answer:
UkoKoshka [18]2 years ago
3 0

Answer:

Magnetic flux, \phi=1.57\times 10^{-5}\ Wb

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,

A=\pi r'^2

A=\pi (0.02)^2=0.00125\ m^2                  

The magnetic flux through this surface is given by :

\phi=B\times A

B is the magnetic field of the solenoid

\phi=\mu_o\dfrac{N}{L}I\times A

\phi=4\pi \times 10^{-7}\times \dfrac{8000}{4}\times 5\times 0.00125

\phi=1.57\times 10^{-5}\ Wb

So, the magnetic flux through this surface is 1.57\times 10^{-5}\ Wb/. Hence, this is the required solution.

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