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?

Answer 1

Good question. I like it.
Do you have any idea?

Answer 2

i also posted here;
but the solution the .sam. got to wasn't even on of the options,
http://openstudy.com/users/unklerhaukus#/updates/516680e6e4b0902dbc1d7bb9
i ended up guessing the wrong solution

it turns out the answer they wanted is \[\frac{\mu_0\pi R}2\]
but i dont really understand

can you explain it ?

Answer 3

B is not uniform on the disc limited by the circuit.
It is maximum at the centre.

So real flux is less than field times area

\(\Phi <B_o \pi R^2=\Large \frac {\mu _o I}{2R}\normalsize \pi R^2=\Large \frac {\mu_o \pi R}{2}\normalsize I\)

Since \(\Phi=L I\) , then \(L>\Large \frac {\mu_o \pi R}{2}\)