Q8P - Fundamentals Of Physics

Question

The figure shows a circular region of radius $\mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{cm}}$in which an electric flux is directed out of the plane of the page. The flux encircled by a concentric circle of radius r is given by ${\mathit{\Phi }}_{\mathbf{E}\mathbf{,}\mathbf{enc}}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{ }\mathbf{\text{Vm/s}}\mathbf{)}\mathbf{(}\mathbf{r}\mathbf{/}\mathbf{R}\mathbf{)}\mathit{t}$, where $\mathit{r}\mathbf{\le }\mathit{R}$and t  is in seconds. (a)What is the magnitude of the induced magnetic field at a radial distance $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{cm}}$? (b)What is the magnitude of the induced magnetic field at a radial distance $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{cm}}$?

Step-by-Step Solution

Answer

1. The magnitude of an induced magnetic field at a radial distance 2 cm is $B=3.54\times {10}^{-17} \mathrm{T}.$
2. The magnitude of an induced magnetic field at a radial distance 5 cm  is  $B=2.13\times {10}^{-17} \mathrm{T}.$

Step 1: Given

$$\begin{gathered} R=3.00 \mathrm{cm} =0.003 \mathrm{m} \\ \mathrm{\varphi }=\left(0.600 \mathrm{V}.\frac{\mathrm{m}}{\mathrm{s}}\right)\frac{\mathrm{r}}{\mathrm{R}}\mathrm{t} \end{gathered}$$

Step 2: Determining the concept

For a non-uniform electric field, we use equation 32-3 for finding the magnetic field inside the circle and outside the circle. If the magnetic induction varies in magnitude and direction at different points in a region, the magnetic field is said to be non-uniform. The magnetic field due to a bar magnet is non-uniform.

The formula is as follows:

$\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}$

Where, $\overrightarrow{B}$ is the magnetic field, $\varphi$ is the flux.  

Step 3: (a) Determining the magnitude of an induced magnetic field at a radial distance 2.00 cm

From the formula, for magnetic field inside the circle, i.e., $r=002$ as

 $\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}$

$\begin{array}{rcl}\oint \overrightarrow{B}·d\overrightarrow{s}& =& {\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}\\ B{\int }_0^{2\pi }rd\theta & =& {\mu }_0{\epsilon }_0\frac{d\varphi }{dt}\\ B2\pi r& =& {\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}\end{array}$

From the given,  $\frac{\varphi }{t}=\left(0.600 \mathrm{V}.\frac{\mathrm{m}}{\mathrm{s}}\right)\frac{r}{R}$.

So, by putting the value,

$\begin{array}{rcl}B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2\pi r}\left(0.600 \right)\frac{r}{R}\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2\pi }\left(0.600 \right)\frac{1}{R}\\ B& =& \frac{\left(2\times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)\left(0.600 \right)}{0.03}\\ B& =& 3.54\times {10}^{-17} \mathrm{T}\end{array}$ 

 

 Therefore, the magnitude of an induced magnetic field at a radial distance 2.00 cm is  $B=3.54\times {10}^{-17} \mathrm{T}.$

Step 4: (b) Determining the magnitude of an induced magnetic field at a radial distance 5.00 cm

As $r > R,$  i.e.,  $r=0.05 \mathrm{m},$ then the $\frac{r}{R}$ is taken as unity, so the equation is

 $\begin{array}{rcl}\oint \overrightarrow{B}·d\overrightarrow{s}& =& {\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}\\ B{\int }_0^{2\pi }rd\theta & =& {\mu }_0{\epsilon }_0\frac{d\varphi }{dt}\\ B2\pi r& =& {\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}\end{array}$

From the given, $\frac{\varphi }{t}=\left(0.600 \mathrm{V}.\frac{\mathrm{m}}{\mathrm{s}}\right)\frac{r}{R}$.

So, by putting the value, 

 $\begin{array}{rcl}B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2\pi r}\left(0.600 \right)\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2\pi r}\left(0.600 \right)\\ B& =& \frac{\left(4\pi \times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)\left(0.600 \right)}{2\pi r}\\ B& =& \frac{\left(2\times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)\left(0.600 \right)}{0.05}\\ B& =& 2.13\times {10}^{-17} \mathrm{T}\\ & & \end{array}$
Therefore, the magnitude of an induced magnetic field at a radial distance $5.00 \mathrm{cm}$ is $B=2.13\times {10}^{-17} \mathrm{T}.$ 

 

By using the concept of magnetic field for the non-uniform electric field, the magnetic field inside and outside the circle can be found.