A circular area with a radius of 6.00 cm lies in the xy-plane. part a what is the magnitude of the magnetic flux through this circle due to a uniform magnetic field with a magnitude of 0.290 t in the + z -direction?
The magnetic flux through an area A is given by [tex]\Phi = BA \cos \theta[/tex] where B is the magnitude of the magnetic field A is the area [tex]\theta[/tex] is the angle between the direction of B and the perpendicular to the surface A.
In our problem, the area lies in the x-y plane, while B is in the z direction, this means that B and the perpendicular to A are parallel, so [tex]\theta=0[/tex] and [tex]\cos \theta=1[/tex], so we can rewrite the formula as [tex]\Phi = BA[/tex]
We can calculate the area starting from the radius: [tex]A= \pi r^2 = \pi (0.06 m)^2 = 0.011 m^2[/tex] And then using the intensity of the magnetic field given by the problem, [tex]B=0.290 T[/tex], we find the magnetic flux: [tex]\Phi = BA=(0.290 T)(0.011 m^2)=3.2 \cdot 10^{-3}Wb[/tex]
