Given data:
Number of turns in the solenoid (N) = 500
Area of cross-section of the solenoid (A) = 2 * 10^-3 m^2
Current passing through the solenoid (I) = 2 A
Magnetic induction (B) = 5 * 10^-3 T

Calculation:
The magnetic flux connected with the solenoid can be calculated using the formula:

Φ = B * A * N

Substituting the given values into the formula:

Φ = (5 * 10^-3 T) * (2 * 10^-3 m^2) * 500

Simplifying the expression:

Φ = 5 * 2 * 10^-3 * 2 * 10^-3 * 500 T * m^2

Φ = 20 * 10^-6 * 1000 T * m^2

Φ = 20 * 10^-3 T * m^2

Φ = 20 * 10^-3 Wb

Therefore, the magnitude of the magnetic flux connected with the solenoid is 20 * 10^-3 Wb.

Explanation:
- A solenoid is a long coil of wire with multiple turns.
- The magnetic flux through a solenoid depends on the number of turns, the area of the cross-section, and the magnetic induction.
- The formula to calculate the magnetic flux is Φ = B * A * N, where Φ is the magnetic flux, B is the magnetic induction, A is the area of cross-section, and N is the number of turns.
- By substituting the given values into the formula and simplifying the expression, we can find the magnitude of the magnetic flux connected with the solenoid.
- In this case, the magnetic flux is calculated to be 20 * 10^-3 Wb.

Summary:
The magnetic flux connected with the solenoid is 20 * 10^-3 Wb. The magnetic flux depends on the number of turns, the area of cross-section, and the magnetic induction. The formula to calculate the magnetic flux is Φ = B * A * N, where Φ is the magnetic flux, B is the magnetic induction, A is the area of cross-section, and N is the number of turns. By substituting the given values into the formula, the magnitude of the magnetic flux connected with the solenoid can be calculated.

Magnetic flux =BAN
=5×10^-3 × 2×10^-3 × 500
=5×10^-3