Two cylindrical solenoids, A and B, each have lengths that are much greater than their diameters. The two solenoids have the same density of turns in their windings. Solenoid A has a diameter that is twice the diameter of solenoid B, but solenoid B has a length that is twice the length of solenoid A. How do their inductances compare with one another

The inductance of solenoid A is twice that of solenoid B

Explanation:

The inductance of a solenoid L is given by

L = μ₀n²Al where n = turns density, A = cross-sectional area of solenoid and l = length of solenoid.

Given that d₁ = 2d₂ and l₂ = 2l₁ and d₁ and d₂ are diameters of solenoids A and B respectively. Also, l₁ and l₂ are lengths of solenoids A and B respectively.

Since we have a cylindrical solenoid, the cross-section is a circle. So, A = πd²/4.

Let L₁ and L₂ be the inductances of solenoids A and B respectively.

So  L₁ = μ₀n²A₁l₁ = μ₀n²πd₁²l₁/4

L₂ = μ₀n²A₂l₂ = μ₀n²πd₂²l₂/4

Since d₁ = 2d₂ and l₂ = 2l₁, sub

L₁/L₂ = μ₀n²πd₁²l₁/4 ÷ μ₀n²πd₂²l₂/4 = d₁²/d₂² × l₁/l₂ = (2d₂)²/d₂² × l₁/2l₁ = 4d₂²/d₂²  × l₁/2l₁ = 4 × 1/2 = 2

L₁/L₂ = 2

L₁ = 2L₂

So, the inductance of solenoid A is twice that of solenoid B

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: