The magnetic flux Φ of a field B on a loop with area A and normal vector n is:
![\Phi=BA\cos\theta](https://img.qammunity.org/2023/formulas/physics/college/esmxw8f70liaejozb7u58tskgirt10rmdn.png)
There θ is the angle between the normal vector n and the field B.
If the normal to the plane of the loop is parallel to the field, then θ=0º. In that case, the magnetic flux is simply the product of the field and the area of the loop, since cos(θ)=0:
![\begin{gathered} \theta=0 \\ \Rightarrow \\ \Phi=BA \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/oc5ba0k66dq7dlxwtsn3etglxtypevho6p.png)
The area of a circle with radius r is:
![A=\pi r^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/lcgfavc89jro4qntamn2b9gfliomu1jwuf.png)
Replace the expression for the area in terms of r and substitute B=1.194T and r=9.792*10^(-2)m to find the magnetic flux:
![\begin{gathered} \Phi=BA \\ \\ =B\cdot\pi r^2 \\ \\ =1.194T\cdot\pi(9.792*10^(-2)m)^2 \\ \\ =0.0359664...Tm^2 \\ \\ \approx0.03597Wb \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/m0a3epepybtm646gkmhq1y3gmvphjsmhwu.png)
Therefore, the magnetic flux if the normal is parallel to the field, is 0.03597 weber.