Let's go step-by-step through both parts of the problem to calculate the magnetic flux.

(a) Magnetic Flux Through the Flat Surface $S_1$:

The magnetic flux $\Phi_B$ through a surface is given by the equation: $\Phi_B = \int_S \mathbf{B} \cdot d\mathbf{A}$ Where:

• $\mathbf{B}$ is the magnetic field vector.
• $d\mathbf{A}$ is a differential area vector on the surface, which points normal (perpendicular) to the surface.

For the flat surface $S_1$, the magnetic flux is easier to calculate because $\mathbf{B}$ is constant and the area vector points directly upward (since the surface is flat). The magnetic field $\mathbf{B}$ is at an angle $\theta$ from the vertical (downward), so it forms an angle $\theta$ with the normal to the surface.

The flux through the flat surface can be calculated as: $\Phi_B = B A \cos(\theta)$ Where:

• $B$ is the magnitude of the magnetic field.
• $A = \pi R^2$ is the area of the flat circular surface, with $R$ being the radius of the hemisphere.
• $\theta$ is the angle between the magnetic field and the vertical direction (which is normal to $S_1$).

Thus, the magnetic flux through $S_1$ is: $\Phi_B = B \pi R^2 \cos(\theta)$

(b) Magnetic Flux Through the Hemispherical Surface $S_2$:

Now let's calculate the flux through the hemispherical surface $S_2$. The magnetic flux through $S_2$ requires us to consider the geometry of the hemisphere and the direction of the magnetic field.

Since the hemisphere is symmetric and $\mathbf{B}$ is at an angle $\theta$ to the vertical, we can still use the magnetic flux formula $\Phi_B = \int_S \mathbf{B} \cdot d\mathbf{A}$. However, the angle between the magnetic field and the area vector changes across the surface of the hemisphere.

For a hemispherical surface of radius $R$, the total area is $2 \pi R^2$. Since the magnetic field is uniform but at an angle $\theta$ to the vertical, the flux through the hemispherical surface is the projection of $\mathbf{B}$ onto the surface area.

The flux through $S_2$ is given by: $\Phi_B = B \cdot \text{Area} \cdot \cos(\theta)$ Where the area of the hemisphere is $2 \pi R^2$.

Thus, the magnetic flux through $S_2$ is: $\Phi_B = B \cdot 2 \pi R^2 \cdot \cos(\theta)$

Final Answers:

• (a) The magnetic flux through the flat surface $S_1$ is: $\Phi_B = B \pi R^2 \cos(\theta)$
• (b) The magnetic flux through the hemispherical surface $S_2$ is: $\Phi_B = 2 B \pi R^2 \cos(\theta)$

Let me know if you'd like further details or if you have any questions!

Additional Questions to Explore:

1. What would happen if the magnetic field were aligned perpendicular to the surface of the hemisphere (i.e., $\theta = 0$)?
2. How would the magnetic flux change if the radius of the hemisphere were doubled?
3. If the magnetic field were to reverse direction, how would the flux through each surface change?
4. How does the magnetic flux through a curved surface compare to that of a flat surface when the same angle $\theta$ is involved?
5. What is the physical significance of the cosine term in the flux equations?

Tip:

Remember that flux depends on the component of the magnetic field that is perpendicular to the surface. The cosine factor accounts for the angle between the magnetic field and the normal to the surface.