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$$\begin{gathered} \stackrel{鈫�}{B_1}=-B_1\hat{k} \\ \stackrel{鈫�}{B_2}=-B_2\hat{k} \\ d\stackrel{鈫�}{A_1}=-d{A}_1\hat{k} \\ d\stackrel{鈫�}{A_2}=-d{A}_2\hat{k} \end{gathered}$$

Radius of circular region $R_1$is $r_1=20\mathrm{cm}=20脳{10}^{-2}m$

Radius of circular region $R_2$is$r_2=30\mathrm{cm}=20脳{10}^{-2}m$

Rate of change of magnetic field isrole="math" localid="1661857590635" $\frac{d{B}_1}{dt}=\frac{d{B}_2}{dt}=8.50脳{10}^{-3}T/s$

The flux of magnetic field through an open surface attached to a closed path is the scalar product of magnetic field and area element of the surface attached to the closed loop. The rate of change of flux is related to the line integral of electric field by Faraday鈥檚 law. So use Faraday鈥檚 law to evaluate the line integral.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

${\stackrel{,}{O}}_{im}=-\frac{d{蠒}_B}{dt}=鈭�\stackrel{鈫�}{E}路d\stackrel{鈫�}{s}$

$d{蠒}_B=\stackrel{鈫�}{B}路d\stackrel{鈫�}{A}$

Where,${桅}_B$is magnetic flux, B is the magnetic field, A is area,饾渶 is emf, E is electric field.

For path 1,

$$\begin{gathered} 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=-\frac{d{蠒}_{B_1}}{dt}=-\frac{d}{dt}\left(鈭�{\stackrel{鈫�}{B}}_1.d\stackrel{鈫�}{A_1}\right)=\frac{d}{dt}\left(-B_1\hat{k}.鈭�d{A}_1\left(-\hat{k}\right)\right) \\ 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=-\frac{d{B}_1}{dt}{A}_1=-蟿蟿r_1^2.\frac{d{B}_1}{dt}=-蟿蟿{\left(20脳{10}^{-2}\right)}^2脳8.50脳{10}^{-3} \\ 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=-3.14脳{\left(0.20\right)}^2脳8.50脳{10}^{-3}=1.073脳{10}^{-3}V \end{gathered}$$

Hence, the electric flux for path 1 is $1.073脳{10}^{-3}V$.

For path 2,

$$\begin{gathered} 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=-\frac{d{蠒}_{B_2}}{dt}=-\frac{d}{dt}\left(鈭�{\stackrel{鈫�}{B}}_2.d\stackrel{鈫�}{A_2}\right)=\frac{d}{dt}\left(+B_2\hat{k}.鈭�d{A}_2\left(\hat{k}\right)\right) \\ 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=-\frac{d{B}_2}{dt}{A}_2=-蟿蟿r_2^2.\frac{d{B}_2}{dt}=-蟿蟿{\left(30脳{10}^{-2}\right)}^2脳8.50脳{10}^{-3} \\ 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=-3.14脳{\left(0.30\right)}^2脳8.50脳{10}^{-3}=-2.40脳{10}^{-3}V \end{gathered}$$

Hence, the electric flux for path 2 is $-2.40脳{10}^{-3}V$.

For path 3,

$$\begin{gathered} 鈭�\stackrel{鈫�}{E}路d\stackrel{鈫�}{S}=-\frac{d{蠒}_{B_3}}{dt} \\ \frac{d{蠒}_{B_3}}{dt}=\frac{d{蠒}_{B_1}}{dt}-\frac{d{蠒}_{B_2}}{dt}=\frac{d}{dt}\left(鈭�{\stackrel{鈫�}{B}}_1.d{\stackrel{鈫�}{A}}_1-鈭�{\stackrel{鈫�}{B}}_2.d{\stackrel{鈫�}{A}}_2\right)=-\frac{d{B}_1}{dt}{A}_1-\left(-\frac{d{B}_2}{dt}{A}_2\right) \\ \frac{d{蠒}_{B_3}}{dt}=\frac{d{B}_1}{dt}{A}_1+\frac{d{B}_2}{dt}{A}_2=-1.073脳{10}^{-3}V+2.40脳{10}^{-3}V=1.327脳{10}^{-3}V \\ 鈭�\stackrel{鈫�}{E}.d\stackrel{鈫�}{s}=\frac{d{蠒}_{B_3}}{dt}=1.327脳{10}^{-3}V \end{gathered}$$

Hence, the electric flux for path 3 is$1.327脳{10}^{-3}V$.

Therefore, by using the concept of flux through magnetic field and Faraday鈥檚 law, the electric flux through a given path can be determined.