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II. Short Answer Questions · Q3

Q.Write down the integral form of modified Ampere's circuital law.

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✓ Free question

Step 1. The original Ampere's circuital law, ∮B⃗⋅dl⃗=μ0iC\oint\vec B\cdot d\vec l=\mu_0 i_C, only accounts for conduction current and fails for surfaces passing through the gap of a charging capacitor.

Step 2. Maxwell added the displacement current term id=ϵ0 dΦE/dti_d=\epsilon_0\,d\Phi_E/dt to fix this, giving the modified law ∮B⃗⋅dl⃗=μ0(iC+id)\oint\vec B\cdot d\vec l=\mu_0(i_C+i_d).

Step 3. Written out explicitly, this is ∮B⃗⋅dl⃗=μ0iC+μ0ϵ0dΦEdt\displaystyle\oint\vec B\cdot d\vec l=\mu_0 i_C+\mu_0\epsilon_0\dfrac{d\Phi_E}{dt}, where iCi_C is the conduction current and ΦE\Phi_E is the electric flux through the surface bounded by the loop.

Step 4. This equation, called the Ampere-Maxwell law, is the fourth of Maxwell's four equations, and reduces to the original Ampere's law whenever the electric flux is constant (steady current).

✓Final answer

∮B⃗⋅dl⃗=μ0iC+μ0ϵ0dΦEdt=μ0(iC+id)\displaystyle\oint\vec B\cdot d\vec l=\mu_0 i_C+\mu_0\epsilon_0\dfrac{d\Phi_E}{dt}=\mu_0(i_C+i_d)

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