Question

Identify the expression of Ampere-Maxwell’s Circuital law.

(a) ∮\( \overset{\rightharpoonup}{B}.\overset{\rightharpoonup}{dl}\) = μ0 (IC-ID)

(b) ∮\( \overset{\rightharpoonup}{B}.\overset{\rightharpoonup}{dl}\) = μ0 (ID-IC)

(c) ∮\( \overset{\rightharpoonup}{B}.\overset{\rightharpoonup}{dl}\) = μ0 (IC+ID)

(d) ∮\( \overset{\rightharpoonup}{B}.\overset{\rightharpoonup}{dl}\) = μ0 (IC ID)

This question was posed to me by my college director while I was bunking the class.

I'm obligated to ask this question of Displacement Current in section Alternating Current of Physics – Class 12

Answer 1

The correct choice is (c) ∮\( \overset{\rightharpoonup}{B}.\overset{\rightharpoonup}{dl}\) = μ0 (IC+ID)

To elaborate: The Ampere-Maxwell Law tells you that this quantity is proportional to the enclosed current and rate of change of electric flux through any surface bounded by the path of integration. The expression for Ampere-Maxwell’s law is given as:

∮\( \overset{\rightharpoonup}{B}.\overset{\rightharpoonup}{dl}\) = μ0(IC+ID)