An equilateral triangle $8.0\mathrm{cm}$on a side is in a $5.0\mathrm{mT}$ uniform magnetic field. The magnetic flux through the triangle is $6.0\mu \mathrm{Wb}$. What is the angle between the magnetic field and an axis perpendicular to the plane of the triangle?

A magnetic flux $\mathrm{\Phi }$seems to be the magnitude of magnetic field which passes through some kind of ring of area $A$. If the magnetic field lies straight towards the plane's normal, This magnetic flux was provided by

${\mathrm{\Phi }}_m=BA$ $\left(1\right)$

The magnetic flux, however, would be supplied by if the magnetic field forms an angular position only with planes.

${\mathrm{\Phi }}_{\mathrm{m}}=BA\cos \theta$ $\left(2\right)$

The entire area of a triangle is equal to half of it's own base times its height. The Pythagoras equation is used to compute the altitude.

$h=\sqrt{(8\mathrm{cm}{)}^2-(4\mathrm{cm}{)}^2}=4\sqrt{3}\mathrm{cm}$

Therefore the area of a triangle,

$A=\frac{1}{2}bh=\frac{1}{2}\left(0.08\mathrm{m}\right)\left(4\sqrt{3}\times {10}^{-2}\mathrm{m}\right)=2.77\times {10}^{-3}{\mathrm{m}}^2$

Then align the equation $\left(2\right)$for localid="1648903666492" $\theta$to obtain

$\theta ={\cos }^{-1}\left(\frac{{\mathrm{\Phi }}_{\mathrm{m}}}{BA}\right)$ (3)

Then put the data for $A$,${\mathrm{\Phi }}_m$and $B$to the equation(3) to obtain role="math" localid="1648903903568" $\theta$

$\theta ={\cos }^{-1}\left(\frac{{\mathrm{\Phi }}_{\mathrm{m}}}{BA}\right)$

$={\cos }^{-1}\left(\frac{6\times {10}^{-6}\mathrm{Wb}}{\left(5\times {10}^{-3}\mathrm{T}\right)\left(2.77\times {10}^{-3}{\mathrm{m}}^2\right)}\right)$

$={64}^{\circ }$