AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Gauss’s law states that, $\phi_{E} = \oint\limits_{s} E dA = \frac{Q}{\varepsilon_0}$ Gauss’s law of magnetism states that the total magnetic flux passing through a bar magnet around its enclosed surface is zero. $\phi_{B} = \oint\limits_{s} B dA = 0$ According to Faraday’s law of electromagnetic induction, $\varepsilon = -\frac{d\phi_{B}}{dt}$ .....(I) The electromotive force in a wire is the line integral, so $\varepsilon = \int E dl$ ....(II) From equation (I) and (II), $\int E dl = -\frac{d\phi_B}{dt}$ In a time varying electric field, there exist an additionalcurrent to the conduction current known as displacement current. So, total current across the loop is, $i = i_c + i_d$ Here, $i_c$ is the conductor current and $i_d$ is the displacement current. Thus, the modified form of the Ampere’s law is, $\oint B dl = \mu_0 i$ $\oint B dl = \mu_0 [i_c + i_d]$ Thus, option (3) is correct.