Answer
a) $\Phi_B = 0$ [Wb]
b) $\Phi_B = -0.0115$ [Wb]
c) $\Phi_B = 0.0115$ [Wb]
d) $\Phi_B = 0$ [Wb]
Work Step by Step
You need to use :
$\vec{B} = 0.128 \vec{a}_z$ [T]
$\Phi_B = \vec{B} \cdot \vec{A} $ where $\vec{A}$ is the surface vector.
a) $\vec{A} = -A \vec{a}_x = -0.12 \vec{a}_x$ [m$^2$]
b) $\vec{A} = -A \vec{a}_z = -0.09 \vec{a}_x$ [m$^2$]
c) $A = 0.15$ [m$^2$]. $\Phi_B = BA\cos(\phi)$ where $\phi$ is the angle between the vector $\vec{B}$ and the vector $\vec{A}$. Here, using trigonometry, you find that $\cos{\phi} = \frac{30}{50}$.
d) Here, $\vec{A}$ is a closed surface. Gauss's second law states that $\oint \vec{B}\cdot \vec{dA} = 0$. So the net flux is null.
