- #1
yungman
- 5,755
- 293
For a coil,
[tex]e=N\frac {d\Phi}{dt}[/tex]
Where [itex]e\;[/itex] is the instantaneous voltage driving the coil and [itex] \Phi\;[/itex] is the flux generated through the coil with N turns.
For a coil
[tex]\oint \vec B \cdot d\vec l =\mu N I \Rightarrow B=\mu N I \Rightarrow \Phi = BS=\mu N I S[/tex]
In the book Handbook of Transformer Design & Application by Flanagan, page 1.7, it gives
[tex]e=N\frac{d\Phi}{dt}\times 10^{-8}[/tex]
It said the multiplier factor depends on the system units. I have no idea how that [itex]10^{-8}\;[/itex] comes from. Please help.
Thanks
Alan
[tex]e=N\frac {d\Phi}{dt}[/tex]
Where [itex]e\;[/itex] is the instantaneous voltage driving the coil and [itex] \Phi\;[/itex] is the flux generated through the coil with N turns.
For a coil
[tex]\oint \vec B \cdot d\vec l =\mu N I \Rightarrow B=\mu N I \Rightarrow \Phi = BS=\mu N I S[/tex]
In the book Handbook of Transformer Design & Application by Flanagan, page 1.7, it gives
[tex]e=N\frac{d\Phi}{dt}\times 10^{-8}[/tex]
It said the multiplier factor depends on the system units. I have no idea how that [itex]10^{-8}\;[/itex] comes from. Please help.
Thanks
Alan