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a solenoid 5 cm in radius and 5 m in length has 8000 uniformly spaced turns and carries a current of 3 a. consider a plane circular surface with radius 2 cm located at the center of the solenoid with its normal axis coincident with the axis of the solenoid. what is the magnetic flux through this surface? (recall that 1

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The magnetic flux through this surface is 7.5 × [tex]10^{-5}[/tex] Wb, for the solenoid with the given radius and length carrying a current of 3 Amperes.

What is Magnetic flux?

The surface integral of the normal component of the magnetic field B over a surface is known as the magnetic flux through that surface. The standard notation is Φ or ΦB. The weber is the SI unit for magnetic flux, and the maxwell is the CGS unit.

Calculation:

It is given that,

Radius of the solenoid, r = 5 cm = 0.05 m

Length of the solenoid, L = 5 m

No of turns, N = 8000

Current, I = 3 A

The radius of the circular plane surface, r' = 2 cm = 0.02

Area of the circular surface,

A = π × r² =  π × (0.02)²= 0.00125 m²

The magnetic flux through this surface is given by :

∅ = B × A

B is the magnetic field of the solenoid,

∅  = μ₀ × N/L × I × A

∅  = 4π × [tex]10^{-7}[/tex] × 8000/5 × 3 × 0.00125 m²

∅  =  7.5 × [tex]10^{-5}[/tex] Wb.

Hence, the magnetic flux through this surface is 7.5 × [tex]10^{-5}[/tex] Wb, for the solenoid with the given radius and length carrying a current of 3 Amperes.

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