91Ó°ÊÓ

The given figure is,

and$\stackrel{→}{B}=\left(5.0m,{30}^{∘}counterclockwisefromvertical\right).$We need to determine the$x$and$y$components of$\stackrel{→}{B}$in coordinate shown in the figure.

The vector $\stackrel{→}{B}=\left(5.0m,{30}^{∘}counterclockwisefromvertical\right)$falls into the second quadrant with the magnitude of $5m.$

$\stackrel{→}{B_x}=\stackrel{→}{B}\sin \mathrm{ϕ}.$

$B_x=5×\sin {30}^{∘}.$

$=5×\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=-2.5m.$

role="math" localid="1649274344760" $\stackrel{→}{B_y}=\stackrel{→}{B}\cos \mathrm{ϕ}.$

$=5×\cos {30}^{∘}.$

$=5×\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

$=\left(4.3m\right)\hat{j}.$

$\stackrel{→}{B}=\left(-2.5m\right)\hat{i}+\left(4.3m\right)\hat{j}.$

The given figure is,

and$\stackrel{→}{B}=\left(5.0m,{30}^{∘}counterclockwisefromvertical\right).$We need to determine the$xandy$components of$\stackrel{→}{B}$in the coordinate system.

The vector $\stackrel{→}{B}=\left(5.0m,{30}^{∘}counterclockwisefromvertical\right).$The $xandy$axis are rotated ${30}^{∘}.\stackrel{→}{B}$perfectly lies in the $y$-axis.

$\mathrm{ϕ}=0^{∘}.$

$\stackrel{→}{B_x}=\stackrel{→}{B}\sin \mathrm{ϕ}.$

$=5m×\sin \left(0^{∘}\right).$

$=\left(0m\right)\hat{i}.$

$\stackrel{→}{B_y}=\stackrel{→}{B}\cos \mathrm{ϕ}.$

$=5×1.$

$=\left(5m\right)\hat{j}.$

$\stackrel{→}{B}=\left(0m\right)\hat{i}+\left(5m\right)\hat{j}.$