Answer :
The magnetic flux through a solenoid of length 35.9 cm, radius 2.5 cm, and 920 turns that carries a current of 1.5 A can be calculated using the formula:
Φ = μ₀ * n * A * I
where Φ represents the magnetic flux, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), n is the number of turns per unit length (in this case, the number of turns divided by the length of the solenoid), A is the cross-sectional area of the solenoid, and I is the current.
To find the cross-sectional area of the solenoid, we can use the formula:
A = π * r²
where A represents the cross-sectional area and r is the radius of the solenoid.
In this case, the radius is given as 2.5 cm, so we have:
A = π * (2.5 cm)²
Now, let's substitute the given values into the formulas:
A = π * (2.5 cm)² = 19.63 cm²
n = 920 turns / 35.9 cm = 25.63 turns/cm
Φ = (4π × 10⁻⁷ T·m/A) * (25.63 turns/cm) * (19.63 cm²) * (1.5 A)
Calculating this expression, we find that the magnetic flux through the solenoid is approximately 13.09 × 10⁻³ Wb (weber).
Therefore, the correct answer is 13.09 times 10⁻³ Wb.
To know more about flux visit:
https://brainly.com/question/29221352
#SPJ11