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The rate of change of flux is given by:

$$\begin{gathered} \frac{d蠒B}{dt}=A\frac{dB}{dt} \\ =蟺r_1^2\frac{dB}{dt}\frac{d蠒B}{dt} \\ =蟺r_1^2\frac{dB}{dt} \end{gathered}$$

Now, according to the faradays law we get:

$$\begin{gathered} 蔚=鈭�\stackrel{鈫�}{E}鈰�\stackrel{炉}{dl} \\ =鈭�\frac{d蠒B}{dt} \end{gathered}$$

Now, the magnetic flux is:$$\begin{gathered} 蠒=AB \\ ={\mathrm{蟺谤}}_1^2\mathrm{B} \end{gathered}$$

Now, integrating LHS gives:

$\begin{array}{r}\stackrel{鈫�}{E}鈰�\stackrel{鈫�}{dl}\\ =E\left(2蟺r_1\right)\end{array}$

Therefore:

$E2蟺r_1=\frac{d\left(B蟺r_1^2\right)}{dt}=蟺r_1^2\frac{dB}{dt}$

Or

$E=\frac{r_1}{2}\frac{dB}{dt}$

The direction of the induced electric field is clockwise.

Since all flux is within r>R then the magnetic field within r_2>R is:

$蠒=AB=蟺R^{2}B$

Integrating LHS we get:

$$\begin{gathered} 鈭�\stackrel{鈫�}{E}鈰�\left(\stackrel{鈫�}{dl}\right)=E\left(2蟺r_2\right) \\ ,E\left(2蟺r_2\right)=蟺R^2\frac{dB}{dt} \end{gathered}$$

So

$E=\frac{R^2}{2r_2}\frac{dB}{dt}$

To draw the graph of the magnitude of the induced electric field as a function as a function of the distance r from the axis $r=0tor=2r$. Is shown below:

To calculate the induced emf in the circular turn radius$r=R/2$. We can use the equation used above with$r1鈫�R/2$

$$\begin{gathered} 蔚=蟺(R/2{)}^2\frac{dB}{dt} \\ =\frac{蟺R^2}{4}\frac{dB}{dt} \\ 蔚=\frac{蟺R^2}{4}\frac{dB}{dt} \end{gathered}$$

Now, the magnitude of the induced emf in a circular turn of radius$r=R.$ we use:

$蔚=\frac{蟺R^2}{4}\frac{dB}{dt}$

Finally, we need to calculate the magnitude of the induced emf in a circular turn of radius$r=2R$. All the flux is within$r<R$, therefore:

$$\begin{gathered} 蔚=\frac{d蠒B}{dt} \\ =蟺R^2\frac{dB}{dt} \end{gathered}$$

Therefore, the induced emf is$蟺R^2\frac{dB}{dt}$.