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Question

# According to the modified Ampere's circuital law (iD= displacement current)

A
Bdl=μ0(iC+ε0dϕEdt)
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B
Bdl=μ0 ε0dϕEdt
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C
Bdl=μ0 iC
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D
Bdl=μ0(iC dϕEdt+iD)
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Solution

## The correct option is A ∮→B⋅→dl=μ0(iC+ε0dϕEdt)According to the modified Ampere's circuital law, ∮→B⋅→dl=μ0(iC+iD)=μ0(iC+ε0dϕEdt) Where, →B→ the magnetic induction at any point on the path. →dl→ is the length element of the path, ic→ is the total conduction current through a surface enclosed by the closed path. Id=ε0dϕdt→ the fictitious displacement current arising from the changing electric flux ϕe. Hence, option (A) is correct.

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