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.

$h=\sqrt{10^2-5^2}=8.66$

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