### Understanding Magnetic Flux in a Solenoid#### Problem Statement**Task:** Calculate the magnetic flux through the surface of a disk-shaped area of radius \( R = 5.00 \, \text{cm} \) that is positioned perpendicular to and centered on the axis of the solenoid.**Given Data:**- Radius of the solenoid (\( r \)) = 1.25 cm- Length of the solenoid (\( L \)) = 26.0 cm- Number of turns = 305- Current (\( I \)) = 20 A#### Diagram ExplanationThe diagram shows a solenoid with:- A cylindrical shape indicated by spiral loops, representing the wire turns.- The length of the solenoid is denoted by \( \ell \).- The radius of the solenoid is marked as \( r \).- A circular area, marked by the blue circle with radius \( R \), is shown perpendicular to the solenoid's axis.#### Calculating Magnetic FluxTo find the magnetic flux (\( \Phi \)) through the disk-shaped area:1. **Magnetic Field Inside a Solenoid:** \[ B = \mu_0 \left( \frac{N}{L} \right) I \] Where: - \( B \) = Magnetic field inside the solenoid - \( \mu_0 \) = Permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)) - \( N \) = Number of turns (305) - \( I \) = Current (20 A) - \( L \) = Length of the solenoid (26.0 cm)2. **Magnetic Flux Calculation:** Since the magnetic field (\( B \)) is uniform inside the solenoid and perpendicular to the disk: \[ \Phi = B \cdot A \] Where: - \( A = \pi R^2 \), the area of the disk (with \( R = 5.00 \, \text{cm} \))#### SolutionThe final solution involves substituting the values into the equations to find the magnetic flux (\( \Phi \)).

Given, radius of the solenoid, r = 1.25 cm,length, L = 26.0 cmnumber of turns, N = 305current, I =…

Answered: A solenoid of radius r = 1.25 cm and…

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A solenoid of radius r = 1.25 cm and length L = 26.0 cm has 305 turns and carries 20 A. Calculate the magnetic flux through the surface of a disk-shapec area of radius R = 5.00 cm that is positioned perpendicular to and centered on the axis of the solenoid. R