Question:
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 $\vec{B}=(3 \hat{i}+4 \hat{k}) \mathrm{T}$. The quantity of flux through the loop $A B C D E F A$ (in $\mathrm{Wb}$ ) is_______
Solution:
Flux through the loop ABCDEFA,
$\phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}) \cdot(25 \hat{\mathrm{i}}+25 \hat{\mathrm{k}})$
$\Rightarrow \phi=(3 \times 25)+(4 \times 25)=175$ weber
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