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  • GIven that <img alt="\varepsilon = \oint \vec E \cdot d\vec l = \frac{- d\phi }{dt}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cvarepsilon%20%3D%20%5Coint%20%5Cvec%20E%20%5Ccdot%2

GIven that \varepsilon = \oint \vec E \cdot d\vec l = \frac{- d\phi }{dt}  Which of the following can be derived from above equation 

Option: 1

\vec V \times \vec E = 0


Option: 2

\vec V \cdot \vec E = -\frac{dB}{dt }


Option: 3

\vec V \cdot \vec E =0


Option: 4

\vec V \times \vec E = -\frac{d\vec B}{dt }


Answers (1)

best_answer

As we have learned

Induced electric field -

\varepsilon = \oint \vec{E }\cdot \vec{dl}= \frac{-d\phi }{dt}

When

A \rightarrow constant

\varepsilon = \oint \vec{E }\cdot \vec{dl}=A\frac{dB}{dt}

B \rightarrow Varying

E= \frac{a^{2}}{2r}\, \frac{dB}{dt}

 

- wherein

E_{inclosed}\, \alpha \frac{1}{r}

a\rightarrow radius

 

 \varepsilon = -\frac{d\phi }{dt}

\oint \vec E \cdot d \vec l =- \int \frac{d \vec B}{dt}\cdot d\vec s          { for a fixed circuit the time derivative will become                                                                          partial time derivative }

Using stoke's theorem 

\int \vec \Delta \times \vec E= - \frac{d \vec B}{dt}

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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