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• Notes

$\begin{array}{l}{\text { A horizontal rectangular surface has dimensions } 2.80 \mathrm{cm}} \\ {\text { by } 3.20 \mathrm{cm} \text { and is in a uniform magnetic field that is directed at an }} \\ {\text { angle of } 30.0^{\circ} \text { above the horizontal. What must the magnitude of }} \\ {\text { the magnetic field be in order to produce a flux of } 4.20 \times 10^{-4} \mathrm{Wb}} \\ {\text { through the surface? }}\end{array}$

$\Phi_{B}=B A \cos \phi$

$B=\frac{\Phi_{B}}{A \cos \phi}=\frac{4 \cdot 2 * 10^{-4} \mathrm{wb}}{(0.028)(0.032) \cos 60^{\circ}}=0.938 \mathrm{T}$

$\begin{array}{l}{\text { An open plastic soda bottle with an opening diameter of }} \\ {2.5 \mathrm{cm} \text { is placed on a table. A uniform } 1.75 \text { -T magnetic field directed }} \\ {\text { upward and oriented } 25^{\circ} \text { from vertical encompasses the bottle. What }} \\ {\text { is the total magnetic flux through the plastic of the soda bottle? }}\end{array}$

$\Phi_{\text { plastic }}+\Phi_{\text { cap }}-0$

$\longrightarrow \quad \Phi_{\text { plastic }}=-\Phi_{\text { cap }}=$$-B A \cos \phi$

$∴ \quad=-B\left(\pi 1^{2}\right) \cos \phi$

$=-(1.75)\left(\pi(0.125)^{2}\right) \cos 25^{0}$

$=\quad -7.8 * 10^{-4} \text { wb }$

$-\Phi_\mathrm{cap }=-7.8 * 10^{-4} \mathrm{wb}=\Phi _\mathrm{plastic}$