To calculate the electric flux through a surface, we utilize the formula for electric flux, which is given by:

\(\Phi_E = E \cdot A \cdot \cos(\theta)\)

In this equation, \(\Phi_E\) represents the electric flux, \(E\) is the magnitude of the electric field, \(A\) is the area of the surface, and \(\theta\) is the angle between the electric field and the normal (perpendicular) vector to the surface.

In this scenario, we know the electric field strength is 100 Newtons per Coulomb and the area of the surface is 1 square meter. The challenge lies in determining the correct angle \(\theta\). It is crucial to note that \(\theta\) should be the angle between the electric field and the normal to the surface, not the angle between the electric field and the surface itself.

Given that the angle between the electric field and the surface is 30 degrees, we can find the angle between the electric field and the normal by recognizing that the normal is perpendicular to the surface. Therefore, the angle we need is:

\(\theta = 90^\circ - 30^\circ = 60^\circ\)

Now, substituting the known values into the electric flux formula:

\(\Phi_E = 100 \, \text{N/C} \cdot 1 \, \text{m}^2 \cdot \cos(60^\circ)\)

Since \(\cos(60^\circ) = \frac{1}{2}\), we can simplify the calculation:

\(\Phi_E = 100 \cdot 1 \cdot \frac{1}{2} = 50 \, \text{N m}^2/\text{C}\)

Thus, the magnitude of the electric flux through the surface is 50 Newton meters squared per Coulomb. This value represents the total electric field passing through the specified area, taking into account the orientation of the surface relative to the electric field.