A loop ABCDEFA of straight edges has six corner points $A(0, 0, 0)$, $B(5, 0, 0)$, $C(5, 5, 0)$, $D(0, 5, 0)$, $E(0, 5, 5)$ and $F(0, 0, 5)$. The magnetic field in this region is $\vec B = (3\hat i + 4\hat k)$ T. The quantity of flux through the loop ABCDEFA (in Wb) is ______.
Numerical value type. Enter your answer.
Answer: 175
Split the loop into two squares by adding the edge DA (traversed both ways, so it adds nothing).
**ABCDA** lies in the $xy$-plane and is traversed anticlockwise seen from $+z$: area vector $25\hat k$.
**ADEFA**: going D → E → F → A in the $yz$-plane, the $(y, z)$ coordinates go $(5,0) \to (5,5) \to (0,5) \to (0,0)$, anticlockwise seen from $+x$: area vector $25\hat i$.
$$\Phi = \vec B\cdot\vec A = (3\hat i + 4\hat k)\cdot(25\hat i + 25\hat k) = 75 + 100 = 175\ \text{Wb}$$
Solution by Sreeraj P, M.Sc Physics