Figure 30-74 shows a uniform magnetic field confined to a cylindrical volume of radius R. The magnitude of is decreasing at a constant rate of 10m Ts. In unit-vector notation, what is the initial acceleration of an electron released at (a) point a (radial distance r=0.05m ), (b) point b (r =0 ), and (c) point c (r =0.05m)?

i) $\frac{\mathrm{dB}}{\mathrm{dt}}=10\mathrm{mT}=0.01\mathrm{T}$

ii) At point a,$r=5\mathrm{cm}=0.05\mathrm{m}$

iii) At point b, r=0

iv) At point c, $r=0.05\mathrm{m}$

By using Faraday’s law, we can find the electric field for the cylinder. After that, we can use the Newton’s law of motion to find out the acceleration in each case.

Formula:

$$\begin{gathered} \int \overrightarrow{E}.d\overrightarrow{s}=-\frac{d\varphi }{dt} \\ {\varphi }_B=BA \end{gathered}$$

According to equation 30-20 which states that change in magnetic field induces an electric field

$$\begin{gathered} \int \overrightarrow{E}.d\overrightarrow{s}=-\frac{d\varphi }{dt} \\ \varphi =BA \end{gathered}$$

For cylinder area $A=2{\mathrm{\pi r}}^2\mathrm{I}$and the electric line are concentrated on the circle having area ${\mathrm{\pi r}}^2$

$$\begin{gathered} E\left(2\mathrm{\pi rI}\right)=-{\mathrm{\pi r}}^2\mathrm{I}\left(\frac{\mathrm{dB}}{\mathrm{dt}}\right) \\ \mathrm{E}=-\frac{{\mathrm{r}}^2}{2}\frac{\mathit{d}\mathit{B}}{\mathit{d}\mathit{t}} \end{gathered}$$

But we know the force on the electron as$\overrightarrow{F}=-e\overrightarrow{E}$

According to Newton’s law, the acceleration is,

$\overrightarrow{a}=-\frac{e\overrightarrow{E}}{m}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

So now at pointelectric field is

$$\begin{gathered} E=-\frac{r}{2}\frac{dB}{dt} \\ E=-\frac{\left(5.0\times {10}^{-2}\right)}{2}\times \left(-10\times {10}^{-3}\right) \\ E=2.5\times {10}^{-4}\mathrm{V}/\mathrm{m} \end{gathered}$$

The magnetic field direction is into the page, so that the direction of electric field at point is to the left, so that $\overrightarrow{E}$direction is

$\overrightarrow{E}=-\left(2.5\times {10}^{-4}\mathrm{V}/\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}$

So that from the equation (1), the acceleration is

$$\begin{gathered} \overrightarrow{a}=-\frac{e\overrightarrow{E}}{m} \\ \overrightarrow{a}=-\frac{e\left(-2.5\times {10}^{-4}V/m\right)}{m}\stackrel{⏜}{i} \end{gathered}$$

By substituting the mass of electron, we get,

$$\begin{gathered} \overrightarrow{a}=-\frac{\left(1.6\times {10}^{-19}\right)\left(-2.5\times {10}^{-4}\mathrm{V}/\mathrm{m}\right)}{9.1\times {10}^{-31}\mathrm{kg}}\stackrel{⏜}{\mathrm{i}} \\ \overrightarrow{\mathrm{a}}=\left(4.4\times {10}^7\mathrm{m}/{\mathrm{s}}^2\right)\stackrel{⏜}{\mathrm{i}} \end{gathered}$$

i.e. the acceleration is to the right.

At point b, the r=0 , so the net acceleration on the charge is also zero.

$\overrightarrow{a}=0$

At point c, the electric field is the same in magnitude as the field in a opposite in direction. So the acceleration

$a_c=-a_a$

So the from resultwe can write as

$\overrightarrow{a}=-\left(4.4\times {10}^{7}m/s^2\right)\stackrel{⏜}{i}$