A solenoid of length 2.8 cm and diameter 0.75 cm is wound with 160 turns per cm. When the current through the solenoid is 0.20 A, what is the magnetic flux through one of the windings of the solenoid?

Question

Quizplus · Accepted Answer

First, we need to find the number of windings in the solenoid:

Number of windings = 160 turns/cm x 2.8 cm = 448

Next, we can use the formula for magnetic flux through a solenoid:

Φ = μ0 * n * A * I

where:
- Φ is the magnetic flux
- μ0 is the permeability of free space (4π x 10^-7 T*m/A)
- n is the number of windings per unit length (160 turns/cm)
- A is the cross-sectional area of the solenoid (πr^2, where r = 0.375 cm)
- I is the current through the solenoid (0.20 A)

Plugging in the values, we get:

Φ = (4π x 10^-7 T*m/A) * (160 turns/cm) * (π x (0.375 cm)^2) * (0.20 A) * 448

Φ = 1.8 x 10^-7 Wb

Therefore, the magnetic flux through one of the windings of the solenoid is 1.8 x 10^-7 Wb, which corresponds to choice B.

A Solenoid of Length 2