yungman
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For a coil,
e=N\frac {d\Phi}{dt}
Where e\; is the instantaneous voltage driving the coil and \Phi\; is the flux generated through the coil with N turns.
For a coil
\oint \vec B \cdot d\vec l =\mu N I \Rightarrow B=\mu N I \Rightarrow \Phi = BS=\mu N I S
In the book Handbook of Transformer Design & Application by Flanagan, page 1.7, it gives
e=N\frac{d\Phi}{dt}\times 10^{-8}
It said the multiplier factor depends on the system units. I have no idea how that 10^{-8}\; comes from. Please help.
Thanks
Alan
e=N\frac {d\Phi}{dt}
Where e\; is the instantaneous voltage driving the coil and \Phi\; is the flux generated through the coil with N turns.
For a coil
\oint \vec B \cdot d\vec l =\mu N I \Rightarrow B=\mu N I \Rightarrow \Phi = BS=\mu N I S
In the book Handbook of Transformer Design & Application by Flanagan, page 1.7, it gives
e=N\frac{d\Phi}{dt}\times 10^{-8}
It said the multiplier factor depends on the system units. I have no idea how that 10^{-8}\; comes from. Please help.
Thanks
Alan