Maxwell's modified form of Ampere's circuital law is

\oint_{s} \to B . \to d s = 0

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\oint_{s} \to B . \to d l = μ_{0} I

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\oint_{s} \to B . \to d l = μ_{0} I + \frac{I}{ε_{o}} \frac{d q}{d t}

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\oint_{s} \to B . \to d l = μ_{0} I + μ_{0} ε_{o} \frac{d ϕ_{E}}{d t}

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Solution

The correct option is D $\oint_{s} \to B . \to d l = μ_{0} I + μ_{0} ε_{o} \frac{d ϕ_{E}}{d t}$
$A n s w e r : -$ D
Treating all charges on the same footing (disregarding whether they are bound or free charges), the generalized Ampère's equation, also called the Maxwell–Ampère equation, is in integral form.

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Maxwell's modified form of Ampere's circuital law is

The correct option is D ∮s→B.→dl=μ0I+μ0εodϕEdtAnswer:− DTreating all charges on the same footing (disregarding whether they are bound or free charges), the generalized Ampère's equation, also called the Maxwell–Ampère equation, is in integral form.