Question

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?

Short answer

The angle between the magnetic field and an axis perpendicular to the plane of the triangle,$\theta ={64}^{\circ }$

Step-by-step solution

The magnetic flux 

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 area of the triangle

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)

The angle θ

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

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

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

$={64}^{\circ }$