Question

Two tightly wound solenoids have the same length and circular cross-sectional area. But solenoid 1 uses wire that is 2.0 times as thick as solenoid 2.

What is the ratio of their inductances?

What is the ratio of their inductive time constants (assuming no other resistance in the circuits)?

Answer 1

a)given d₁ =2d₂ =>d₁/d₂ =2

Inductance of a solenoid
L =?_oAN²/l

Because they are tightly wound, the number of turns is determined by the diameter of the wire:

N =l/d

If we form the ratio of inductances for the two conditions, we have

L₁/L₂ =(N₁/N₂)² =(d₂/d₁)² =(1/2)^2=0.25

b)The length of wire used for the turns is

l_wire =N(?D)

where D =diameter of th esolenoid.

Thus for the ratio of resistances, we have

R₁/R₂ = (l_wire1/l_wire2)(d₂/d₁)² = (N₁/N₂)(d₂/d₁)² = (d₂/d₁)³

For the ratio of the time constants, we get

?₁/?₂ = (L₁/L₂)(R₂/R₁) = (L₁/L₂)(d₁/d₂)³ =(0.25)(2)^3 =2

I hope help you.....