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 magnetic field inside the solenoid using the formula:
B = μ₀nI
where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.

μ₀ = 4π×10⁻⁷ T m A⁻¹ (permeability of free space)
n = 160 turns/cm = 1600 turns/m (number of turns per unit length)
I = 0.20 A (current)

B = (4π×10⁻⁷ T m A⁻¹) x (1600 turns/m) x (0.20 A) = 4.03 x 10⁻³ T

Now, we can find the magnetic flux through one of the windings using the formula:
Φ = BA
where Φ is the magnetic flux, B is the magnetic field, and A is the area of one winding.

A = πr² = π(0.375 cm)² = 0.44 cm² = 4.4 x 10⁻⁵ m² (area of one winding)
B = 4.03 x 10⁻³ T (magnetic field)

Φ = (4.03 x 10⁻³ T) x (4.4 x 10⁻⁵ m²) = 1.77 x 10⁻⁷ Wb

Therefore, the magnetic flux through one of the windings of the solenoid is 1.8 x 10⁻⁷ Wb, which is closest to choice (B).

A Solenoid of Length 2 ×\times× 10-7 Wb B) 1 ×\times× 10-7 Wb C) 0

A Solenoid of Length 2 $\times$ 10^-7 Wb
B) 1 $\times$ 10^-7 Wb
C) 0