Intuition The one-line ideaA changing electric field acts like a current — it produces a magnetic field just as a flowing charge does. Maxwell added this "current" to fix a contradiction in the original Ampère's law, and in doing so predicted electromagnetic waves .
The original Ampère's law says: the magnetic field circulating around a closed loop equals the current threading any surface bounded by that loop.
∮ B ⃗ ⋅ d l ⃗ = μ 0 I e n c \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc} ∮ B ⋅ d l = μ 0 I e n c
Intuition Why "any surface"?A closed loop is just a wire ring. You can stretch infinitely many soap-film surfaces across it like a bubble. Ampère's law must give the same answer for all of them — otherwise the law is meaningless.
The charging capacitor paradox. Consider a wire charging a parallel-plate capacitor.
Surface 1 (flat disc cutting the wire): the conduction current I I I ∮ B ⃗ ⋅ d l ⃗ = μ 0 I \oint\vec B\cdot d\vec l = \mu_0 I ∮ B ⋅ d l = μ 0 I Surface 2 (bulging surface passing between the plates): no charge flows across the gap → I e n c = 0 I_{enc}=0 I e n c = 0 ∮ B ⃗ ⋅ d l ⃗ = 0 \oint\vec B\cdot d\vec l = 0 ∮ B ⋅ d l = 0
Same loop, two answers. Contradiction. Something is missing in the gap.
Common mistake "Charge jumps the gap, so current still flows" — why this feels right, and the fixIt feels right because the wire current is continuous outside. But no charge crosses the vacuum gap — electrons pile on one plate, leave the other. The real fix isn't moving charge; it's that the growing electric field between the plates plays the role of a current. (Steel-man corrected.)
Definition Displacement currentThe quantity I d = ε 0 d Φ E d t I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} I d = ε 0 d t d Φ E Φ E = ∫ E ⃗ ⋅ d A ⃗ \Phi_E=\int\vec E\cdot d\vec A Φ E = ∫ E ⋅ d A not a flow of charge — it is the magnetic-field-producing effect of a changing electric flux .
Intuition The deep WHY — symmetry → wavesFaraday: a changing B ⃗ \vec B B E ⃗ \vec E E E ⃗ \vec E E B ⃗ \vec B B feed each other — a self-sustaining ripple that travels at c = 1 / μ 0 ε 0 c=1/\sqrt{\mu_0\varepsilon_0} c = 1/ μ 0 ε 0 I d I_d I d no light .
Worked example 1. B-field inside a charging capacitorCircular plates radius R = 5 R=5 R = 5 I = 2 I=2 I = 2 B B B r = 3 r=3 r = 3
Step 1 — symmetry: B B B r r r ∮ B ⃗ ⋅ d l ⃗ = B ( 2 π r ) \oint\vec B\cdot d\vec l = B(2\pi r) ∮ B ⋅ d l = B ( 2 π r ) Why? Same symmetry as a wire.
Step 2 — displacement current enclosed by radius r r r I d I_d I d I d , e n c = I π r 2 π R 2 = I r 2 R 2 I_{d,enc}=I\frac{\pi r^2}{\pi R^2}=I\frac{r^2}{R^2} I d , e n c = I π R 2 π r 2 = I R 2 r 2 Why? Flux through the small loop is the fraction r 2 / R 2 r^2/R^2 r 2 / R 2 Step 3 — apply law: B ( 2 π r ) = μ 0 I r 2 R 2 B(2\pi r)=\mu_0 I\dfrac{r^2}{R^2} B ( 2 π r ) = μ 0 I R 2 r 2 B = μ 0 I r 2 π R 2 = ( 4 π × 10 − 7 ) ( 2 ) ( 0.03 ) 2 π ( 0.05 ) 2 = 4.8 × 10 − 6 T B=\frac{\mu_0 I r}{2\pi R^2}=\frac{(4\pi\times10^{-7})(2)(0.03)}{2\pi(0.05)^2}=4.8\times10^{-6}\,\text{T} B = 2 π R 2 μ 0 I r = 2 π ( 0.05 ) 2 ( 4 π × 1 0 − 7 ) ( 2 ) ( 0.03 ) = 4.8 × 1 0 − 6 T Why this form? Identical to B B B
Worked example 2. NumericalI d I_d I d C = 2 μ F C=2\,\mu\text{F} C = 2 μ F d V / d t = 10 6 dV/dt=10^6 d V / d t = 1 0 6 I d I_d I d
Step 1 — Q = C V ⇒ I d = d Q d t = C d V d t Q=CV \Rightarrow I_d=\dfrac{dQ}{dt}=C\dfrac{dV}{dt} Q = C V ⇒ I d = d t d Q = C d t d V Why? We proved I d = I = d Q / d t I_d=I=dQ/dt I d = I = d Q / d t Step 2 — I d = ( 2 × 10 − 6 ) ( 10 6 ) = 2 I_d=(2\times10^{-6})(10^6)=2 I d = ( 2 × 1 0 − 6 ) ( 1 0 6 ) = 2 Why it matters: You never measured the field directly, yet got the displacement current — it always equals the conduction current feeding the cap.
Worked example 3. From flux directlyElectric flux through a region grows at d Φ E / d t = 4 × 10 6 V⋅m/s d\Phi_E/dt=4\times10^{6}\ \text{V·m/s} d Φ E / d t = 4 × 1 0 6 V⋅m/s I d I_d I d
I d = ε 0 d Φ E d t = ( 8.85 × 10 − 12 ) ( 4 × 10 6 ) = 3.5 × 10 − 5 I_d=\varepsilon_0\dfrac{d\Phi_E}{dt}=(8.85\times10^{-12})(4\times10^6)=3.5\times10^{-5} I d = ε 0 d t d Φ E = ( 8.85 × 1 0 − 12 ) ( 4 × 1 0 6 ) = 3.5 × 1 0 − 5 Why? Direct definition — no charges needed at all (this works in pure vacuum waves).
Recall Feynman: explain to a 12-year-oldImagine two metal plates with a gap. You push electricity into one plate. No spark crosses the gap, yet a compass near the gap still twitches — a magnet field appears! Why? Because the invisible electric field in the gap is growing , and Maxwell realised a growing electric field behaves exactly like a real electric current. So the magnetic field "doesn't notice" the gap. This trick is why light exists: an electric wiggle makes a magnetic wiggle makes an electric wiggle... forever, racing across space.
"Changing E is the secret current."
I d = ε 0 Φ ˙ E I_d = \varepsilon_0 \dot\Phi_E I d = ε 0 Φ ˙ E Epsilon-zero times flux-dot ."
Faraday: moving B → E . Maxwell: moving E → B . The mirror twins.
Why does the original Ampère's law fail at a capacitor gap?
Does displacement current involve moving charge?
Why must I d I_d I d I c I_c I c
What constant emerges from μ 0 ε 0 \mu_0\varepsilon_0 μ 0 ε 0
What flaw in Ampère's law did Maxwell fix? For a charging capacitor, two surfaces on the same loop give different
I e n c I_{enc} I e n c (wire vs. gap), a contradiction.
Define displacement current. I d = ε 0 d Φ E / d t I_d=\varepsilon_0\,d\Phi_E/dt I d = ε 0 d Φ E / d t , the magnetic effect of a changing electric flux (not moving charge).
Why must I d = I c I_d = I_c I d = I c Charge conservation:
I = d Q / d t I=dQ/dt I = d Q / d t and
Φ E = Q / ε 0 \Phi_E=Q/\varepsilon_0 Φ E = Q / ε 0 give
ε 0 d Φ E / d t = d Q / d t = I \varepsilon_0\,d\Phi_E/dt = dQ/dt = I ε 0 d Φ E / d t = d Q / d t = I .
State the Ampère–Maxwell law. ∮ B ⃗ ⋅ d l ⃗ = μ 0 I c + μ 0 ε 0 d Φ E / d t \oint\vec B\cdot d\vec l=\mu_0 I_c+\mu_0\varepsilon_0\,d\Phi_E/dt ∮ B ⋅ d l = μ 0 I c + μ 0 ε 0 d Φ E / d t .
What does displacement current predict physically? Self-sustaining EM waves travelling at
c = 1 / μ 0 ε 0 c=1/\sqrt{\mu_0\varepsilon_0} c = 1/ μ 0 ε 0 .
Does a spark/charge cross the capacitor gap? No — only the electric field changes; that change plays the role of current.
B inside a charging circular capacitor at radius r < R r<R r < R B = μ 0 I r / ( 2 π R 2 ) B=\mu_0 I r/(2\pi R^2) B = μ 0 I r / ( 2 π R 2 ) .
I d I_d I d I d = C d V / d t I_d=C\,dV/dt I d = C d V / d t .
Charging capacitor paradox
Derived from charge conservation
Intuition Hinglish mein samjhoDekho, original Ampère's law kehta hai ki magnetic field ka loop integral μ 0 \mu_0 μ 0 I I I
Maxwell ne genius move kiya: bola ki gap mein bhale charge na chal raha ho, par electric field to badal raha hai (kyunki plate par charge jama ho raha hai). Yeh changing electric field bhi ek "current" ki tarah behave karta hai — isko bola displacement current , I d = ε 0 d Φ E / d t I_d=\varepsilon_0\,d\Phi_E/dt I d = ε 0 d Φ E / d t I d I_d I d I c I_c I c
Iska matlab kyun important hai? Faraday ne dikhaya tha changing B B B E E E E E E B B B c = 1 / μ 0 ε 0 c=1/\sqrt{\mu_0\varepsilon_0} c = 1/ μ 0 ε 0 light khud Maxwell ke is addition ka result hai! Bina displacement current ke, electromagnetic waves exist hi nahi karte.
Exam tip: yaad rakho I d = C d V / d t I_d=C\,dV/dt I d = C d V / d t B = μ 0 I r / ( 2 π R 2 ) B=\mu_0 I r/(2\pi R^2) B = μ 0 I r / ( 2 π R 2 )
