Added Luis's English solution of 11.2.2

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+ <title>The induction of a uniform magnetic field inside a cylinder of radius $r$ = 0.1 m increases linearly with time: $B = \alpha t$ (coefficient $\alpha$ = 10$^{-3}$ T/s). The magnetic field is directed along the axis of the cylinder. What is the strength of the eddy electric field at a distance of $l$ = 0.2 m from the cylinder axis?</title>
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+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#11.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $11.2.2$
+ The induction of a uniform magnetic field inside a cylinder of radius $r$ = 0.1 m increases linearly with time: $B = \alpha t$ (coefficient $\alpha$ = 10$^{-3}$ T/s). The magnetic field is directed along the axis of the cylinder. What is the strength of the eddy electric field at a distance of $l$ = 0.2 m from the cylinder axis?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ Magnetic field passes through an area of $\pi r^2$ m$^2$ and its flux increases with time. So, this causes that appears a rotational electric field around cylinder axis such that its induced magnetic field opposes to $\vec{B}(t)$ (Lenz Law). Applying Faraday's Law,
+ $$\oint\vec{E}\cdot \vec{ds} = \frac{d\Phi_B}{dt}$$
+ this closed integral is for concentric circular paths about cylinder's axis. Then, for a distance $a$ from center,
+ $$E(a) \cdot 2\pi a = \pi r^2 \alpha$$
+ $$E(a) = \frac{\alpha r^2}{2a}$$
+ Finally, evaluating for $a=l$,
+ $$E = \frac{\alpha r^2}{2l}$$
+ </p>
+ <h4>Answer</h4>
+ <p>
+ $$E = 2.5~\cdot~10^{-5}~{\rm{\frac{V}{m}}}$$
+ </p>
+
+
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
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