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What is the flux, in milliWebers, through the triangle?

User Tom Seddon
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1 Answer

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Based on the drawing, half of the area has a flux going into the plane and other going outside the plane.


h=√(10^2-5^2)=8.66


\begin{gathered} \phi=B* S \\ \phi=(B1S)/(2)+(B2S)/(2)=(B1+B2)(S)/(2) \\ \phi=(0.3-0.1)(10\cdot8.66)(1)/(2\cdot2) \\ \phi=0.2T\cdot21.65cm^2 \\ \phi=4.33Tcm^2 \end{gathered}
\phi=4.33\cdot10^(-1)mWb

What is the flux, in milliWebers, through the triangle?-example-1
User Nathan Liang
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