91Ó°ÊÓ

1. Radius of region 1,${\mathrm{r}}_1=1.0\mathrm{cm}=1×{10}^{-2}\mathrm{m}$
2. Radius of region 2,${\mathrm{r}}_2=2.0\mathrm{cm}=2×{10}^{-2}\mathrm{m}$
3. Rate of change of magnetic field in region 1,i.e.$\frac{{\mathrm{dB}}_1}{\mathrm{dt}}$ , is increasing in magnitude
4. Vertical scale,${\mathrm{E}}_{\mathrm{s}}=20\mathrm{nV}$

We know from the Faraday’s law that the changing magnetic field can induce an emf. Therefore, the rate of change of magnetic field is equal to the induced emf. By using the data from the given plot, we can find the rate of change of magnetic field in both the regions. We can find if the magnitude of $\stackrel{\mathbf{→}}{{\mathbf{B}}_{\mathbf{2}}}$is increasing or decreasing by considering the Lenz’s law.

Formula:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{ind}}=-\frac{{\mathrm{»åÏ•}}_{\mathrm{B}}}{\mathrm{dt}} \\ {\mathrm{Ï•}}_{\mathrm{B}}=\mathrm{Ï•}\stackrel{→}{\mathrm{B}}.\mathrm{d}\stackrel{→}{\mathrm{A}} \end{gathered}$$

Let’s find the slope $m_1$from the graph, showing variation of E verses ${\mathrm{R}}^2$in the region 1

role="math" localid="1661856139421" ${\mathrm{m}}_1=\frac{\mathrm{E}}{{\mathrm{R}}^2}=\frac{8.0-0.0}{1.0-0.0}.\frac{\mathrm{nV}}{{\mathrm{cm}}^2}=8.0\frac{\mathrm{nV}}{{\mathrm{cm}}^2}$

Similarly, forthe region 2, we get

${\mathrm{m}}_2=\frac{\mathrm{E}}{{\mathrm{R}}^2}=\frac{20.0-8.0}{4.0-1.0}.\frac{\mathrm{nV}}{{\mathrm{cm}}^2}=\frac{12}{3}\frac{\mathrm{nV}}{{\mathrm{cm}}^2}=4.0\frac{\mathrm{nV}}{{\mathrm{cm}}^2}$

Now from Faraday’s law, induced emf is given as

${\mathrm{E}}_{\mathrm{ind}}=-\frac{{\mathrm{»åÏ•}}_{\mathrm{B}}}{\mathrm{dt}}=-\mathrm{A}.\frac{\mathrm{dB}}{\mathrm{dt}}$

Taking the magnitude, we have

$\left|{\mathrm{E}}_{\mathrm{ind}}\right|=\left|\frac{{\mathrm{»åÏ•}}_{\mathrm{B}}}{\mathrm{dt}}\right|=-\mathrm{A}.\frac{\mathrm{dB}}{\mathrm{dt}}$

Where A is the area attached to the loop along which we intend to find the emf.

For the region1, we have,

$E=\mathrm{Ï€}{r}_1^2.\frac{d{B}_1}{dt}$

Using the above expression, we get,

$$\begin{gathered} \frac{\mathrm{E}}{{\mathrm{r}}_1^2}=\mathrm{Ï€}.\frac{{\mathrm{dB}}_1}{\mathrm{dt}} \\ \frac{{\mathrm{dB}}_1}{\mathrm{dt}}=\frac{{\mathrm{m}}_1}{\mathrm{Ï€}}=\frac{\stackrel{Ë™}{\mathrm{o}}}{{\mathrm{r}}_1^2}=\frac{1}{\mathrm{Ï€}}=\frac{8.0}{\mathrm{Ï€}}\frac{\mathrm{nV}}{{\mathrm{cm}}^2}=\frac{80}{\mathrm{Ï€}}\frac{\mathrm{μ³Õ}}{{\mathrm{m}}^2}≈25\mathrm{μ°Õ}/\mathrm{s} \end{gathered}$$

F the region 2, we have

Using above expression, we get,

$$\begin{gathered} \frac{\mathrm{E}}{{\mathrm{r}}_2^2}=\mathrm{Ï€}.\frac{{\mathrm{dB}}_2}{\mathrm{dt}} \\ \frac{{\mathrm{dB}}_2}{\mathrm{dt}}=\frac{{\mathrm{m}}_2}{\mathrm{Ï€}}=\frac{\mathrm{E}}{{\mathrm{r}}_2^2}.\frac{1}{\mathrm{Ï€}}=\frac{4.0}{\mathrm{Ï€}}\frac{\mathrm{nV}}{{\mathrm{cm}}^2}=\frac{4.0}{\mathrm{Ï€}}\frac{\mathrm{μ³Õ}}{{\mathrm{m}}^2}≈13\mathrm{μ°Õ}/\mathrm{s} \end{gathered}$$

From the slope of the graph, the showing variation of E verses $R^2$ , for region 2, the induced emf is increasing, therefore by Lenz’s law we can say that the magnitude of$\stackrel{→}{B_2}$ is increasing.