# Proof that $\text{Magnetic Flux Density} = {\text{Magentic Flux} \over A\sin(θ)}?$

Problem : How is $$\text{Magnetic Flux Density} = {\text{Magnetic Flux} \over A\sin(θ)}$$ (this is the formula my text book has given me)

Where $$θ$$ is the angle between magnetic flux lines and plane and $$A$$ is the Area

I know that $$\text{Magnetic Flux Density} = {\text{Magnetic Flux} \over A}$$ but where does that $$\sin(θ)$$ come from ? Why is the formula multiplied by $$1 \over \sin(θ)$$ ?

• The element of flux is $|A| {\bf n}\cdot {\bf B}$ where $|A|$ is the area and ${\bf n}$ is the unit normal to the surface. This is $|A| \cos \theta$. For some reason your book uses $90-\theta$ hence the $\sin\theta$. – mike stone Sep 22 at 21:40
• Is is not just the definition of the dot product? ${\bf n}\cdot {\bf B}= |{\bf n}| |{\bf B}| \cos \theta$? Geometrically its just that it's only when the area is face-on that you see all of it. If it's at an angle you only see a fraction $\cos \theta$ of it. Edge on, there is no area for the field to penentrate. – mike stone Sep 22 at 23:32