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

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