- #1
lelouch_v1
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I have been reading Griffith's Introduction to Electodynamics and i am currently at the chapter about magnetostatics. There is an example about a long solenoid with n units per length and radius R that shows a way of finding the magnetic vector potential. The magnetic field inside the solenoid is
$$\textbf{B} = {\mu_0}nI{\hat{z}} \ \ , \ \ {\text{inside solenoid}}$$ and $$\textbf{B} = 0 \ \ , \ \ {\text{outside solenoid}}$$
What i cannot understand is that when dealing with the magnetic vector potential, Griffiths states that magnetic flux outside of the solenoid is
$$\int{\textbf{B}{\cdot}d{\textbf{a}}} = {\mu_0}nI({\pi}R^2)$$ and
$$\textbf{A}=\frac{\mu_0nIR^2}{2s}{\hat{\phi}} \ \ , \ \ s>=R$$
Why do we use the magnetic field that we have on the inside, when trying to find the magnetic flux on the outside?
$$\textbf{B} = {\mu_0}nI{\hat{z}} \ \ , \ \ {\text{inside solenoid}}$$ and $$\textbf{B} = 0 \ \ , \ \ {\text{outside solenoid}}$$
What i cannot understand is that when dealing with the magnetic vector potential, Griffiths states that magnetic flux outside of the solenoid is
$$\int{\textbf{B}{\cdot}d{\textbf{a}}} = {\mu_0}nI({\pi}R^2)$$ and
$$\textbf{A}=\frac{\mu_0nIR^2}{2s}{\hat{\phi}} \ \ , \ \ s>=R$$
Why do we use the magnetic field that we have on the inside, when trying to find the magnetic flux on the outside?