Question

Figure shows two circular regions R₁ and R₂ with radii ${\mathit{r}}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{r}}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$. In R₁ there is a uniform magnetic field of magnitude${\mathit{B}}_{\mathbf{1}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{T}$ directed into the page, and in R₂ there is a uniform magnetic field of magnitude ${\mathit{B}}_{\mathbf{1}}\mathbf{=}\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{T}$directed out of the page (ignore fringing). Both fields are decreasing at the rate of 8.50 mT/s. (a) Calculate $\mathbf{\oint }\overrightarrow{\mathbf{E}}\mathbf{.}\overrightarrow{\mathbf{d}\mathbf{s}}$for path 1.(b) Calculate $\mathbf{\oint }\overrightarrow{\mathbf{E}}\mathbf{.}\overrightarrow{\mathbf{ds}}$for path 2.(c) Calculate $\mathbf{\oint }\overrightarrow{\mathbf{E}}\mathbf{.}\overrightarrow{\mathbf{ds}}$for path 3.

Short answer

1. The electric flux for path 1 is $-1.073\times {10}^{-3}V$.

b. The electric flux for path 2 is $-2.40\times {10}^{-3}V$.

c. The electric flux for path 3 is $1.327\times {10}^{-3}V$.

Step-by-step solution

Given

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

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

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

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

Determining the concept

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’s law. So use Faraday’s 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:

${\stackrel{,}{O}}_{im}=-\frac{d{\varphi }_B}{dt}=\oint \overrightarrow{E}·d\overrightarrow{s}$

$d{\varphi }_B=\overrightarrow{B}·d\overrightarrow{A}$

Where,${\Phi }_B$is magnetic flux, B is the magnetic field, A is area,𝜀 is emf, E is electric field.

(a) Determining the electric flux for path 1

For path 1,

$$\begin{gathered} \oint \overrightarrow{E}.d\overrightarrow{s}=-\frac{d{\varphi }_{B_1}}{dt}=-\frac{d}{dt}\left(\oint {\overrightarrow{B}}_1.d\overrightarrow{A_1}\right)=\frac{d}{dt}\left(-B_1\hat{k}.\oint d{A}_1\left(-\hat{k}\right)\right) \\ \oint \overrightarrow{E}.d\overrightarrow{s}=-\frac{d{B}_1}{dt}{A}_1=-\tau \tau r_1^2.\frac{d{B}_1}{dt}=-\tau \tau {\left(20\times {10}^{-2}\right)}^2\times 8.50\times {10}^{-3} \\ \oint \overrightarrow{E}.d\overrightarrow{s}=-3.14\times {\left(0.20\right)}^2\times 8.50\times {10}^{-3}=1.073\times {10}^{-3}V \end{gathered}$$

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

(b) Determining the electric flux for path 2

For path 2,

$$\begin{gathered} \oint \overrightarrow{E}.d\overrightarrow{s}=-\frac{d{\varphi }_{B_2}}{dt}=-\frac{d}{dt}\left(\oint {\overrightarrow{B}}_2.d\overrightarrow{A_2}\right)=\frac{d}{dt}\left(+B_2\hat{k}.\oint d{A}_2\left(\hat{k}\right)\right) \\ \oint \overrightarrow{E}.d\overrightarrow{s}=-\frac{d{B}_2}{dt}{A}_2=-\tau \tau r_2^2.\frac{d{B}_2}{dt}=-\tau \tau {\left(30\times {10}^{-2}\right)}^2\times 8.50\times {10}^{-3} \\ \oint \overrightarrow{E}.d\overrightarrow{s}=-3.14\times {\left(0.30\right)}^2\times 8.50\times {10}^{-3}=-2.40\times {10}^{-3}V \end{gathered}$$

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

(c) Determining the electric flux for path 3

For path 3,

$$\begin{gathered} \oint \overrightarrow{E}·d\overrightarrow{S}=-\frac{d{\varphi }_{B_3}}{dt} \\ \frac{d{\varphi }_{B_3}}{dt}=\frac{d{\varphi }_{B_1}}{dt}-\frac{d{\varphi }_{B_2}}{dt}=\frac{d}{dt}\left(\oint {\overrightarrow{B}}_1.d{\overrightarrow{A}}_1-\oint {\overrightarrow{B}}_2.d{\overrightarrow{A}}_2\right)=-\frac{d{B}_1}{dt}{A}_1-\left(-\frac{d{B}_2}{dt}{A}_2\right) \\ \frac{d{\varphi }_{B_3}}{dt}=\frac{d{B}_1}{dt}{A}_1+\frac{d{B}_2}{dt}{A}_2=-1.073\times {10}^{-3}V+2.40\times {10}^{-3}V=1.327\times {10}^{-3}V \\ \oint \overrightarrow{E}.d\overrightarrow{s}=\frac{d{\varphi }_{B_3}}{dt}=1.327\times {10}^{-3}V \end{gathered}$$

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

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