Question

QUESTION 3

We have seen that the magnetic field at the center of a circular loop of radius \(R\) of current \(I\) is \(B=\frac{\mu_0I}{2R}\). What is a lower bound estimate of the inductance of this single loop of wire?

its multiple a choice,
im not sure what calculation they want me to do?

Answer 1

@.Sam.

Answer 2

multiple choice?

Answer 3

\[•\frac{\mu_0}{2R}\]\[•{\mu_0}{\pi R}\]\[•\frac{\mu_0\pi R^2}2\]or\[•\frac{\mu_0\pi R}2\]

Answer 4

any idea?

Answer 5

i'm thinking it's •\(\frac{\mu_0\pi R^2}2\)
but i can't be sure

Answer 6

how come

Answer 7

I have no clue ,Only since he didn't mention N (number of turns) that's mean its only one turn which is directly prop. to the B at the center of a circular loop ..
I would go with the third choice but not sure too :/

Answer 8

Sry I fell asleep just now... They already said that its single loop, so if N=1, \[\oint B • dl= \mu_OI \\ \\ B(2\pi)(R)=\mu_OI \\ \\ B=(\frac{\mu_O}{2R})\frac{I}{\pi}\]---------------------------------------\[L=\frac{\Phi N}{i} \\ \\ L=\frac{BAN}{i} \\ \text{Subbing B in,} \\ L=\frac{(\frac{I}{\pi})(\frac{\mu_O}{2R})(A)(N)}{i}\]\[L=\frac{\mu_O \cancel I}{2\pi R}(\frac{\cancel N^1A}{\cancel i}) \\ \\ L=\frac{\mu_OR}{2\pi}\]Hmm...