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$$
\begin{array}{l}{\text { In a physics laboratory experiment, a coil with } 200 \text { turns }} \\ {\text { enclosing an area of } 12 \mathrm{cm}^{2} \text { is rotated in } 0.040 \text { s from a position }} \\ {\text { where its plane is perpendicular to the earth's magnetic field to a }} \\ {\text { position where its plane is parallel to the field. The earth's mag- - }} \\ {\text { netic field at the lab location is } 6.0 \times 10^{-5} \mathrm{T} \text { . (a) What is the total }} \\ {\text { magnetic flux through the coil before it is rotated? After it is }} \\ {\text { rotated? (b) What is the average emf induced in the coil? }}\end{array}
$$

$$
|\varepsilon|=\left|\frac{d \Phi_{B}}{d t}\right|, \phi_{B}=B A \cos \phi
$$

$$
\phi_{i^{\circ}}=o^{\circ}
\quad ,$$$$
\phi_{f}=9{0}^{\circ}
$$

(a) $$
\phi_{B i^{\circ}}=B A \cos o^{\circ}
$$$$
=6 * 10^{-5} *\left(12*10^{-4}\right)(1)=7 \cdot 2*10^{-8} \mathrm{wb}
$$

$$
N \Phi_{B i}=
$$$$
200 * 7 \cdot 2*10^{-8}=1.44 * 10^{-5} \mathrm{wb}
$$

$$
\Phi_{B F}=B A \cos 90^{\circ}=0
$$

$$
\left|\mathcal{E}_{\text { averg }}\right|
$$$$
=\left|\frac{N \Phi_{i^\circ}-N \Phi_{f}}{\Delta t}\right|=
$$$$
\left|\frac{1.44 * 10^{-5}-0}{0.04}\right|=
$$$$
3.6 * 10^{-4} \mathrm{V}
$$

$$
=0.36 \mathrm{mV}
$$

$$
\begin{array}{l}{\text { A coil } 4.00 \mathrm{cm} \text { in radius, containing } 500 \text { turns, is }} \\ {\text { placed in a uniform magnetic field that varies with time according }} \\ {\text { to } B=(0.0120 \mathrm{T} / \mathrm{s}) t+\left(3.00 \times 10^{-5} \mathrm{T} / \mathrm{s}^{4}\right) t^{4} . \text { The coil is con- }} \\ {\text { nected to a } 600-\Omega \text { resistor, and its plane is perpendicular to the }} \\ {\text { magnetic field. You can ignore the resistance of the coil. (a) Find }} \\ {\text { the magnitude of the induced emf in the coil as a function of time. }} \\ {\text { (b) What is the current in the resistor at time } t=5.00 \mathrm{s} ?}\end{array}
$$

$$
|\varepsilon|=\frac{N d \Phi_{B}}{d t}
$$$$
=N A \frac{d B}{d t}
$$$$
=N A \frac{d}{d t}(0.012 t*3*10^{-5} t^{4}
$$

(a) $$
\rightarrow|\varepsilon|=N A
$$$$
\left(0.012+1.2*10^{-4} t^{3}\right)
$$$$
=500 *\left(\pi(0.04)^{2}\right)\left[0.012+1.2*10^{-4} t^{3}\right]
$$

$$
=(0.0302 \mathrm{V})+\left(3.02 * 10^{-4} \mathrm{V} / \mathrm{s}^{3}\right) \mathrm{t}^{3}
$$

(b) $$
I=\frac{\varepsilon}{R}
$$ (a) $$
t=5 s \rightarrow|\varepsilon|
$$$$
=0.0302+\left(3.02*10^{-4}\right)(5)^{3}=0.068 \mathrm{V}
$$

$$
\rightarrow \quad I=\frac{0.068}{600}=1.13 * 10^{-4} \mathrm{A}
$$

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