Question

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

Answer 1

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`