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find 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. 8.73 times 10-3 wb 13.09 times 10-3 wb 349.04 times 10-3 wb 87.26 times 10-3 wb 2.18 times 10-3 wb

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 = μâ‚€ * n * A * I

where Φ represents the magnetic flux, μâ‚€ 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.

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