Question

A 0.5 T magnetic field is normal to the plane of a circular loop with a radius of 0.25 m. What is the magnetic flux through the loop?

Answer 1

\[\Phi=B·S\]\[B=0.5T\]\[S=\pi R^2\]\[\Phi=0.5·\pi ·0.25^2T \times m^2\]
where 1 T x m^2= 1 Weber

Answer 2

Magnetic flux is the amount of magnetic field lines which passes perpendicular to a plane.

We define a plane as a vector that is normal to the plane of the surface. The reason is it has more practical uses then defining it other ways.

\[\Phi_{B}=\vec B\cdot \vec A = BAcos\theta\] where theta is the angle between the normal of the plane and the magnetic field.

Here, A is the area of the circular loop and B is the magnetic field strength of 0.5T. The angle is perpendicular to the circular loop, and the vector that describes this circular loop is also perpendicular to the loop, so the angle θ is 0.

And so the problem quickly reduces to:
\[\Phi_{B}=0.5·(\pi ·0.25^2)\]