Time constant of a Toroidal Solenoid

[Abo]

Homework Statement

It says that you have a Toroidal Solenoid that has a certain self inductance L and an inner resistance (self resistance) R. So it has a time constant (tawo). Now if you double the number of loops of this solenoid, without changing anything else, what will the time constant be?

Relevant Equations

Φ = B * A
B = (μ0 * N * I) /2 π r
L = N (Φ / I)

Is my solution reasonable?
What I got from my first attempt is that the time constant won't change. WHY? Because when we double the number of loops (N) we're going to have new values for both the self inductance and the resistance of the solenoid and so the ratio (L/R) stays the same. Here is a photo of my solution .

 

[kuruman]

Check your derivation of $R_2=4R_1$. What is the equation for the resistance in terms of the resistivity, length and cross sectional area of the wire? What changes and what does not when you double the number of loops?

 

[Abo]

resistance = resistivity * length / area
well it is obvious that the resistance will increase when using a longer wire.

length = N * (μ0 * I)/B
is it right to say that"double N" gives us "double length" here?
if so it is, then R2 = 2R1
and the time constant will be two times longer.
Am I right this time?

 

[kuruman]

You are right. The time constant is $\tau=L/R$. $L$ increases by a factor of 4, $R$ increases by a factor of 2, therefore the ratio increases by a factor of $4/2=2$.

 

[Abo]

Thank you for taking the time to answer! I appreciate it!

 

[kuruman]

You are welcome.