Question

Solve for the magnetic flux through a loop with an area of 8.5 ✕ 10 -4 m2 found perpendicular to a magnetic field with magnitude 0.75 T.

Answer 1

For the flat surface the magnetic flux is equal to $\Phi = \vec{B}\cdot\vec{S} = BS\cos \theta,$ where $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the surface.

In our case the magnetic field is perpendicular to the surface, so $\theta = 0,\; \cos\theta = 0$ and $\Phi = BS = 0.75\,\mathrm{T}\cdot 8.5\cdot10^{-4}\,\mathrm{m^2} = 6.375\cdot10^{-4}\,\mathrm{Wb}$ .