Question

The correct form of Ampere's law for circuits with gaps in them is (a) B.ds o (b) fB.ds=lenclosed (c) fB-ds-μ0|enclosed. Fig. 1 (d) fB.ds=Halenclosed+A4쓸 (e) B ds-do 0 enclosed

Answer 1

this is the basic circuital law(ignore the double integral, it means for integrating over the surface )

${\displaystyle \oint _{C}\mathbf {B} \cdot \mathrm {d} {\boldsymbol {l}}=\mu _{0}\iint _{S}\mathbf {J} \cdot \mathrm {d} \mathbf {S} =\mu _{0}I_{\mathrm {enc} }}$

but for the above case, there is polarization density for the given boundary .then the basic circuit law can be modified with maxwell ampere equation

then we get (integral form )

${\displaystyle \oint _{C}\mathbf {B} \cdot \mathrm {d} {\boldsymbol {l}}=\iint _{S}\left(\mu _{0}\mathbf {J} +\mu _{0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\right)\cdot \mathrm {d} \mathbf {S} }$

or in the differential form, it is

${\displaystyle \mathbf {\nabla } \times \mathbf {B} =\mu _{0}\mathbf {J} +\mu _{0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}}$ so the answer is D