$$
\begin{array}{l}{\text { A metal ring } 4.50 \mathrm{cm} \text { in diameter is placed between the }} \\ {\text { north and south poles of large magnets with the plane of its area }} \\ {\text { perpendicular to the magnetic field. These magnets produce an ini- - }} \\ {\text { tial uniform field of } 1.12 \text { T between them but are gradually pulled }} \\ {\text { apart, causing this field to remain uniform but decrease steadily at }}\end{array}
$$

$$
\begin{array}{l}{0.250 \mathrm{T} / \mathrm{s} . \text { (a) What is the magnitude of the electric field induced }} \\ {\text { in the ring? (b) In which direction (clockwise or counterclockwise) }} \\ {\text { does the current flow as viewed by someone on the south pole of }} \\ {\text { the magnet? }}\end{array}
$$

$$
\begin{array}{l}{\text { A long, straight solenoid with a cross-sectional area of }} \\ {8.00 \mathrm{cm}^{2} \text { is wound with } 90 \text { turns of wire per centimeter, and the }} \\ {\text { windings carry a current of } 0.350 \text { A. A second winding of } 12 \text { turns }} \\ {\text { encircles the solenoid at its center. The current in the solenoid is }} \\ {\text { turned off such that the magnetic field of the solenoid becomes }} \\ {\text { zero in } 0.0400 \text { s. What is the average induced emf in the second }} \\ {\text { winding? }}\end{array}
$$

$$

\begin{array}{l}{\text { A metal ring } 4.50 \mathrm{cm} \text { in diameter is placed between the }} \\ {\text { north and south poles of large magnets with the plane of its area }} \\ {\text { perpendicular to the magnetic field. These magnets produce an ini- - }} \\ {\text { tial uniform field of } 1.12 \text { T between them but are gradually pulled }} \\ {\text { apart, causing this field to remain uniform but decrease steadily at }}\end{array}

$$

$$

\begin{array}{l}{0.250 \mathrm{T} / \mathrm{s} . \text { (a) What is the magnitude of the electric field induced }} \\ {\text { in the ring? (b) In which direction (clockwise or counterclockwise) }} \\ {\text { does the current flow as viewed by someone on the south pole of }} \\ {\text { the magnet? }}\end{array}

$$

$$

\oint \vec{E} \cdot \overrightarrow{d L}=-\frac{d \Phi_{\mathrm{B}}}{d t}

$$

$$

E 2 \pi r =

$$$$

\frac{d(B \pi r)^{2} }{d t}

$$$$

=\pi r^{2} \frac{d B}{d t}

$$

$$\rightarrow

E=\frac{r}{2} \frac{d B}{d t}

$$

\(∴ E=\frac{0.0225 \mathrm{m}}{2}(0.25)=2.81 * 10^{-3} \mathrm{V} / \mathrm{m} \)

$$

\begin{array}{l}{\text { A long, straight solenoid with a cross-sectional area of }} \\ {8.00 \mathrm{cm}^{2} \text { is wound with } 90 \text { turns of wire per centimeter, and the }} \\ {\text { windings carry a current of } 0.350 \text { A. A second winding of } 12 \text { turns }} \\ {\text { encircles the solenoid at its center. The current in the solenoid is }} \\ {\text { turned off such that the magnetic field of the solenoid becomes }} \\ {\text { zero in } 0.0400 \text { s. What is the average induced emf in the second }} \\ {\text { winding? }}\end{array}

$$

$$

\left|\varepsilon_{avrg }\right|=

$$$$

N\left|\frac{\Delta \Phi_{B}}{\Delta t}\right|

$$

$$

B=M_{0} nI

$$

$$

\Phi_{B}=B A

$$

$$

\rightarrow\left|\varepsilon_{avrg}\right|=

$$$$

N\left|\frac{\Delta \Phi_{B}}{\Delta t}\right|

$$$$

=| \frac{N A \Delta B}{\Delta t} )=

$$$$

\frac{N A M_0n I}{\Delta t}

$$

$$

\rightarrow\varepsilon_{avrg}

$$$$=

\frac{4 \pi*10^{-7} * 12 *\left(8 * 10^{-4}\right)\left(9000 m^{-1}\right)(0.35)}{0.04}

$$$$

=9.5 * 10^{-4} \mathrm{V}

$$

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