In short

Ampère's circuital law (for steady currents) states: the line integral of the magnetic field \vec{B} around any closed loop equals \mu_0 times the total steady current passing through any surface bounded by the loop.

\boxed{\;\oint\vec{B}\cdot d\vec{\ell} \;=\; \mu_0\,I_\text{enc}\;}

Here d\vec{\ell} is an infinitesimal directed element of the closed Amperian loop; I_\text{enc} is the net current piercing any surface capped on that loop (with sign determined by the right-hand rule applied to the loop's circulation direction); and \mu_0 = 4\pi\times 10^{-7}\ \text{T}\cdot\text{m}/\text{A} is the permeability of free space.

Ampère is to Biot-Savart what Gauss is to Coulomb. Both Gauss and Ampère replace volume/line integrals of the source (charge, current) with surface/line integrals of the field — provided the geometry has enough symmetry for the field magnitude to be pulled out of the integral. When that happens, you get the field in three lines instead of pages of integration.

How to use it — the four-step recipe.

  1. Identify the symmetry of the current distribution (cylindrical, planar, toroidal).
  2. Deduce the direction and symmetry of \vec{B} on purely geometric grounds.
  3. Choose an Amperian loop on which either \vec{B} is constant and parallel to d\vec{\ell} (so \vec{B}\cdot d\vec{\ell} = B\,d\ell) or perpendicular to d\vec{\ell} (so \vec{B}\cdot d\vec{\ell} = 0). The magnitude of B must be constant along every part of the loop where it is parallel.
  4. Compute I_\text{enc} (careful with signs and with current-density integrals), equate to \mu_0 I_\text{enc}, and solve for B.

Classic results derived in two-line calculations.

  • Infinite straight wire: B = \mu_0 I/(2\pi r)
  • Solenoid interior: B = \mu_0 n I (covered in the next chapter)
  • Toroid interior: B = \mu_0 N I/(2\pi r)

Limitation and Maxwell's fix. Ampère's law in the form above holds only for steady currents (currents constant in time). For time-varying currents, it fails — and the repair is Maxwell's displacement current I_d = \varepsilon_0\,d\Phi_E/dt:

\oint\vec{B}\cdot d\vec{\ell} \;=\; \mu_0\,(I_\text{enc} + I_d).

This correction is what turned four separate laws into Maxwell's equations and predicted electromagnetic waves travelling at the speed of light.

Stand at the edge of a 400 kV transmission line corridor — the towers you pass on the Mumbai-Pune expressway, carrying power from Tarapur Atomic Power Station to the Maharashtra grid. A current of perhaps 800 A flows through the three bundled conductors strung between the towers. Ask: what is the magnetic field two metres below the lowest conductor? The Biot-Savart law gives you the answer if you are willing to integrate, but the cable runs for kilometres in both directions and the integral is painful. There has to be a better way.

Here is the better way. The current distribution is symmetric — infinitely long, straight, cylindrical. The only magnetic-field geometry consistent with this symmetry is a field that circles the wire at each point, with a magnitude that depends only on distance from the wire (not on which point along the wire you are under). Draw a circle of radius r = 2 m centred on the wire, passing through the point you care about. The field is tangent to this circle and constant in magnitude on it. Walk once around the circle:

\oint\vec{B}\cdot d\vec{\ell} \;=\; B\cdot 2\pi r.

Whatever current is enclosed by the circle — in this case the full 800 A of the line — must appear on the right side of Ampère's law with the universal factor \mu_0:

B\cdot 2\pi r \;=\; \mu_0 I \quad\Longrightarrow\quad B \;=\; \frac{\mu_0 I}{2\pi r} \;=\; \frac{4\pi\times 10^{-7}\times 800}{2\pi\times 2} \;=\; 8\times 10^{-5}\ \text{T}.

That is 1.6 times Earth's field, 2 m from a Maharashtra-grid transmission line. Three lines of algebra. No integration. That is what Ampère's circuital law does.

This chapter states the law carefully, shows you how to pick an Amperian loop that exploits the symmetry, derives three classic results that every JEE Advanced student should be able to reproduce on a blank sheet, explains why the law fails for time-varying currents, and introduces Maxwell's displacement-current fix that completed the equations of electromagnetism in 1864.

The statement of Ampère's law

Take any closed curve C in space. This curve is an Amperian loop — it is a mathematical object, not a physical wire. Walk once around C in a chosen direction, integrating \vec{B}\cdot d\vec{\ell} as you go. The result is

\oint_C \vec{B}\cdot d\vec{\ell} \;=\; \mu_0\,I_\text{enc} \tag{1}

where I_\text{enc} is the net current piercing any surface bounded by C, with the sign convention that currents flowing in the direction given by the right-hand rule applied to the loop's circulation direction count as positive.

Two things to notice immediately.

First, the law does not care about the shape of the Amperian loop, or about the details of where the enclosed current flows. It only cares about (i) the loop's total circulation direction and (ii) the net current threading any capping surface. Any two different choices of capping surface (a flat disc, a hemispherical cap, a bag-shaped surface) give the same I_\text{enc} because current is conserved: whatever flows in must flow out through another face, so the net flow through any two surfaces with the same boundary must be equal.

Second, the law is a global statement. It says the sum of \vec{B}\cdot d\vec{\ell} around a loop is determined by the enclosed current — not that the \vec{B} on a single piece of the loop comes only from the enclosed current. Every current in the universe (enclosed or not) contributes to \vec{B} at every point on the loop. The magic is that when you integrate around the closed loop, the contributions from non-enclosed currents integrate to zero, and only the enclosed currents leave a residue.

An Amperian loop with two enclosed and one external currentA closed curve labeled C in the plane of the page, with arrows indicating counterclockwise circulation direction. Three straight currents pierce the plane: two upward arrows labeled I1 and I2 inside the loop, and one downward arrow labeled I3 outside the loop. A right-hand indicates that the loop's circulation direction corresponds to positive current coming out of the page. Amperian loop C I₁ (out) I₂ (out) I₃ (in, external) ∮ B · dℓ = μ₀(I₁ + I₂) — I₃ does not contribute
An Amperian loop $C$ (red dashed) with three currents piercing the plane of the page. $I_1$ and $I_2$ are enclosed (out of page = positive by the right-hand rule applied to the counterclockwise loop circulation); $I_3$ pierces the plane outside the loop and does not contribute to $I_\text{enc}$. The law says $\oint_C\vec{B}\cdot d\vec{\ell} = \mu_0(I_1 + I_2)$, with every current in space (including $I_3$) contributing to $\vec{B}$ at individual points but only the enclosed currents summing up in the loop integral.

Ampère versus Gauss — the same idea, one dimension different

If you know Gauss's law for electrostatics,

\oint\vec{E}\cdot d\vec{A} \;=\; \frac{Q_\text{enc}}{\varepsilon_0},

then Ampère's law should look familiar. Both say: a surface/line integral of the field equals the enclosed source, times a universal constant. The only structural difference is dimensional:

Gauss Ampère
Source charge (scalar) current (flux of current density)
Field electric \vec{E} (radial from charges) magnetic \vec{B} (circling currents)
Integral \oint\vec{E}\cdot d\vec{A} over closed surface \oint\vec{B}\cdot d\vec{\ell} around closed loop
Works when spherical / cylindrical / planar charge symmetry cylindrical / planar / toroidal current symmetry
Universal constant 1/\varepsilon_0 \mu_0

Gauss's law is always true; it is only useful when the charge distribution has enough symmetry that you can pull E out of the surface integral. Ampère's law is always true for steady currents; it is only useful when the current distribution has enough symmetry that you can pull B out of the line integral. In both cases the "trick" is to find the right closed path on which the integrand is either constant (so you can factor it out) or zero (so it doesn't contribute).

Choosing an Amperian loop

This is the hard and interesting part. Before you can use equation (1), you must pick a loop C on which \oint\vec{B}\cdot d\vec{\ell} simplifies to something you can compute — ideally B\cdot L where L is the length of a "good" portion of the loop and B is constant on it. Three symmetry patterns cover almost every JEE Advanced problem.

Cylindrical symmetry — straight wires, coaxial cables, solid cylinders with current. Use a circle centred on the symmetry axis. On the circle, \vec{B} is tangent (by symmetry, no other direction is invariant under rotation about the axis), and |\vec{B}| is constant. So \oint\vec{B}\cdot d\vec{\ell} = B\cdot 2\pi r.

Planar symmetry — infinite current sheet. Use a rectangular loop with two sides parallel to the current direction (above and below the sheet) and two sides perpendicular to it. On the parallel sides \vec{B}\cdot d\vec{\ell} = \pm B\cdot L; on the perpendicular sides \vec{B}\cdot d\vec{\ell} = 0 (because \vec{B} is tangent to the sheet and the path segments are perpendicular to the sheet's plane direction, or by symmetry the contribution cancels). So \oint\vec{B}\cdot d\vec{\ell} = 2BL.

Toroidal symmetry — solenoids and toroids. Use a rectangular loop partly inside and partly outside the coil (for a solenoid), or a circle coaxial with the toroid. On the "good" portion where \vec{B} is parallel to d\vec{\ell} and constant, \oint\vec{B}\cdot d\vec{\ell} simplifies; on the other portions it is zero or cancels. Details in the next chapter.

The recipe is: let the symmetry of the current dictate the shape of the loop. Never guess.

Explore the enclosed current

The interactive figure below shows a solid cylindrical conductor of radius a carrying total current I uniformly distributed over its cross-section. For any Amperian loop (a circle of radius r coaxial with the conductor), the enclosed current varies with r: for r < a, only a fraction of the total pierces the loop; for r > a, all of I does. Drag r and watch both I_\text{enc}(r) and the resulting B(r) = \mu_0 I_\text{enc}/(2\pi r) evolve.

Interactive: magnetic field inside and outside a cylindrical conductor A plot of B(r), the magnetic field at radial distance r from the axis of a long solid cylindrical conductor of radius 0.02 m carrying a total uniformly-distributed current of 50 A. The field rises linearly from zero at the centre to its peak at r = 0.02 m, then falls as 1 over r outside the conductor. A draggable dot selects r and readouts give I_enc and B. radial distance r (m) B(r) (T) 0 250 µT 500 µT 0 0.02 0.04 0.08 r = a (surface) B(r) drag the red dot
A solid cylindrical conductor of radius $a = 0.02$ m carrying $I = 50$ A uniformly over its cross-section. For $r < a$, $I_\text{enc}(r) = I\cdot (r^2/a^2)$ and $B(r) = \mu_0 I r/(2\pi a^2)$ — linearly rising. For $r > a$, $I_\text{enc} = I$ (all current enclosed) and $B(r) = \mu_0 I/(2\pi r)$ — the infinite-wire result, falling as $1/r$. Peak at $r = a$ (the surface of the conductor).

Worked examples

Example 1: Infinite straight wire — Ampère in three lines

A very long straight wire carries steady current I. Find the magnetic field at a perpendicular distance r from the wire.

Amperian loop around a long straight wireA long vertical wire carrying current I upward. An Amperian loop that is a circle of radius r, centered on the wire, is drawn in the horizontal plane perpendicular to the wire. The magnetic field B at each point of the circle is tangent to it, in the direction given by the right-hand rule. The integral of B around the circle equals mu_0 times I. I (up) loop C (radius r) r B (tangent) B B B
A long straight wire carrying current $I$ up the page. The chosen Amperian loop is a circle of radius $r$ perpendicular to the wire, centred on it. Four sample points on the circle show $\vec{B}$ tangent to the circle, in the direction of the right-hand curl. Magnitude of $\vec{B}$ is the same at every point on the circle by symmetry.

Step 1. Use symmetry to determine \vec{B}.

The current distribution is invariant under:

  • translation along the wire (an infinite wire looks the same if you slide it up);
  • rotation about the wire (the wire looks the same from any angle around it);
  • reflection in any plane containing the wire.

\vec{B} must share these symmetries. Translation invariance means \vec{B} does not depend on z (the position along the wire). Rotation invariance means |\vec{B}| depends only on the perpendicular distance r. Reflection in a plane containing the wire reverses the component of \vec{B} along the radial direction (if any), so that component must be zero — no radial \vec{B}.

Reflection in the plane perpendicular to the wire would reverse the current direction, and also would reverse B_z (the component along the wire). The current does reverse under this reflection, so we cannot use the reflection to kill B_z directly. But we can use the curl of the right-hand rule: every current element d\vec{l} = \hat{z}\,dz has d\vec{l}\times\hat{r} in the tangent direction \hat{\phi}, never in the \hat{z} direction. So B_z = 0 too.

Conclusion: \vec{B} = B(r)\,\hat{\phi} — tangent to circles around the wire, magnitude dependent only on r.

Why: nine-tenths of Ampère's-law problems are solved by symmetry. Before you ever write down the law, you must be able to say: "the field points this way, and depends on this coordinate." If you cannot, the geometry does not have enough symmetry for Ampère to be useful, and you should use Biot-Savart instead.

Step 2. Choose the Amperian loop.

Pick the circle of radius r centred on the wire, lying in a plane perpendicular to the wire, with circulation direction consistent with the right-hand rule (thumb along I, fingers curl in the positive direction).

Why: this is the loop on which \vec{B} and d\vec{\ell} are parallel at every point, and |\vec{B}| is constant. Both properties are essential for pulling B out of the integral.

Step 3. Compute the line integral.

\oint\vec{B}\cdot d\vec{\ell} \;=\; \oint B(r)\,d\ell \;=\; B(r)\oint d\ell \;=\; B(r)\cdot 2\pi r.

Step 4. Compute the enclosed current.

The Amperian circle threads the wire once, and the right-hand rule for the chosen circulation direction gives I_\text{enc} = +I.

Step 5. Equate and solve.

B(r)\cdot 2\pi r \;=\; \mu_0\,I \quad\Longrightarrow\quad \boxed{\;B(r) \;=\; \frac{\mu_0\,I}{2\pi\,r}.\;} \tag{2}

Result: B = \mu_0 I/(2\pi r), tangent to circles around the wire, direction given by the right-hand rule.

What this shows: The Biot-Savart derivation of the same result (Example 2 of the previous chapter) was a trigonometric-substitution integral. Ampère gives the same answer in three lines of symmetry argument plus three lines of arithmetic. That is the power of a symmetry-based law — when the geometry allows it, the work is done.

Example 2: Coaxial cable — inside, between, and outside

A coaxial cable has an inner solid conductor of radius a carrying current I up the page, and an outer cylindrical shell at radius b (thickness negligible) carrying the same current I down the page. Find B(r) in three regions: (i) r < a, (ii) a < r < b, (iii) r > b.

Coaxial cable cross-sectionA cross-sectional view of a coaxial cable: inner conductor of radius a carrying current out of the page, shown with a dot; outer thin shell at radius b carrying current into the page, shown with crosses around the circumference. Three concentric circles represent Amperian loops: one at r less than a, one between a and b, one at r greater than b. outer shell, radius b b I (out) a I (into page) r < a a < r < b r > b
Cross-section of the coaxial cable viewed with the inner current coming *out* of the page (dot) and the outer return current going *into* the page (crosses). Three Amperian circles (dashed red) — one of each region — are used to find $B(r)$ in each zone.

Step 1. Symmetry argument (identical to Example 1).

By cylindrical symmetry and invariance along the axis, \vec{B} = B(r)\,\hat{\phi} in every region. Pick an Amperian circle of radius r centred on the axis, with circulation matching the right-hand rule for current up the page (counterclockwise when viewed from above).

Step 2. Region (i): r < a (inside the inner conductor).

Assume the current is uniformly distributed over the inner conductor's cross-section (standard JEE assumption). Then the current density is

J \;=\; \frac{I}{\pi a^2},

and the current enclosed by the Amperian circle is the fraction (r/a)^2 of the total:

I_\text{enc} \;=\; J\cdot\pi r^2 \;=\; I\cdot\frac{r^2}{a^2}.

Why: I_\text{enc} is the current passing through the disc bounded by the Amperian circle. Uniform density means the current is proportional to the disc's area, which is \pi r^2; dividing by the conductor's cross-section \pi a^2 gives the fraction.

Apply Ampère:

B(r)\cdot 2\pi r \;=\; \mu_0\cdot I\cdot\frac{r^2}{a^2} \quad\Longrightarrow\quad \boxed{\;B(r) \;=\; \frac{\mu_0 I r}{2\pi a^2}\quad(r < a).\;}

The field grows linearly inside the conductor, from 0 at the centre to \mu_0 I/(2\pi a) at the surface.

Step 3. Region (ii): a < r < b (between inner and outer conductors).

Now the full inner current +I is enclosed, and the outer shell is outside the loop. So I_\text{enc} = I.

B(r)\cdot 2\pi r \;=\; \mu_0 I \quad\Longrightarrow\quad \boxed{\;B(r) \;=\; \frac{\mu_0 I}{2\pi r}\quad(a < r < b).\;}

Same form as the infinite-wire result — the field in the annular gap decays as 1/r.

Step 4. Region (iii): r > b (outside the cable).

Now both the inner current +I (out of page) and the outer current -I (into page) are enclosed:

I_\text{enc} \;=\; +I + (-I) \;=\; 0.
B(r)\cdot 2\pi r \;=\; 0 \quad\Longrightarrow\quad \boxed{\;B(r) \;=\; 0\quad(r > b).\;}

Outside the cable, there is no magnetic field at all.

Result: B(r) grows linearly inside the inner conductor, falls as 1/r in the gap, and is exactly zero outside.

What this shows: This is why coaxial cables carry video and data signals in every Indian television system (from the Doordarshan era to today's cable TV) with almost no electromagnetic leakage. The outer conductor is not decoration — it is the return path that cancels the magnetic field outside the cable. No leaked \vec{B} means no radiated power, no interference with nearby circuits, and no crosstalk with the cables running alongside. The engineering-electromagnetics argument for coaxial geometry is exactly this Ampère's-law calculation.

Example 3: Infinite current sheet

An infinite thin sheet of current carries surface current density \vec{K} (amperes per metre of width, in the direction of current flow). Find the magnetic field on both sides.

Infinite current sheet with rectangular Amperian loopAn infinite horizontal current sheet in the plane of the page, with surface current K flowing to the right shown as arrows along the sheet. A rectangular Amperian loop straddles the sheet with its top side of length L above the sheet and its bottom side of length L below, and vertical sides connecting them. B field arrows point in opposite directions on the two sides of the sheet, parallel to the top and bottom segments. sheet of current K (out of the edge) Amperian loop L B (above, rightward) B (below, leftward)
An infinite current sheet with surface current $K$ (amperes per unit width) flowing to the right. By symmetry, $\vec{B}$ is uniform in magnitude on each side, tangent to the sheet, and parallel to it but perpendicular to $\vec{K}$. Above the sheet $\vec{B}$ points in one direction; below, the opposite. The Amperian rectangle spans both sides with horizontal segments of length $L$ parallel to $\vec{B}$.

Step 1. Symmetry.

Translational symmetry in both directions in the plane of the sheet means |\vec{B}| depends only on the perpendicular distance from the sheet, not on position along the sheet. Reflection in the plane of the sheet reverses the current direction, so \vec{B} must be parallel to the sheet (no perpendicular component). A reflection that swaps K's direction must flip \vec{B}'s direction; so if you are above the sheet and look down at \vec{K} pointing right, \vec{B} (on your side) points perpendicular to \vec{K} in the sheet's plane — specifically, opposite to the direction on the other side.

Step 2. Amperian loop.

Choose a rectangular loop straddling the sheet, with its top segment of length L above the sheet at some height, bottom segment of length L below at the same distance, and two vertical segments connecting them. Orient the rectangle so the top and bottom segments lie along the direction of \vec{B} (perpendicular to \vec{K} in the sheet's plane).

Step 3. Line integral.

On the top segment, \vec{B}\cdot d\vec{\ell} = +BL (parallel). On the bottom segment, \vec{B} has reversed direction but d\vec{\ell} has also reversed (the loop walks the other way on the bottom segment), so we get +BL again. On the vertical segments, \vec{B} is perpendicular to d\vec{\ell} (since \vec{B} lies in the plane of the sheet, parallel to the horizontal segments), so \vec{B}\cdot d\vec{\ell} = 0.

\oint\vec{B}\cdot d\vec{\ell} \;=\; BL + BL + 0 + 0 \;=\; 2BL.

Step 4. Enclosed current.

The loop has width L along the current direction, so the current piercing any capping surface is I_\text{enc} = KL (surface current density times length).

Step 5. Apply Ampère.

2BL \;=\; \mu_0\,KL \quad\Longrightarrow\quad \boxed{\;B \;=\; \frac{\mu_0\,K}{2}\;}

Notice: the field is independent of distance from the sheet. Unlike a point source whose field falls off with distance, an infinite sheet produces a uniform field everywhere on each side. This is analogous to the uniform electric field of an infinite charged plane — both are the "simplest" field shapes in their respective subjects.

Result: B = \mu_0 K/2 on each side of an infinite current sheet, directions opposite, magnitude independent of distance.

What this shows: No physical current sheet is infinite, but the result is a good approximation for large, thin, uniformly-conducting sheets. It is also the building-block result used to derive the solenoid field (by treating the solenoid's wall as a stack of current sheets). And it is the reason a single finite conducting sheet shields modestly well at short range but not at far — you would need an enclosure (like the Faraday shield around a transformer) to kill the field completely.

Common confusions

If you came here to use Ampère's law for JEE Advanced cylindrical / planar / toroidal problems, you have what you need. What follows is the history of why the law needed correction, Maxwell's displacement-current fix, and how that correction predicted the existence of electromagnetic waves.

The flaw — a parallel-plate capacitor charging up

Consider a parallel-plate capacitor being charged through a wire carrying current I. Outside the capacitor, the current flows through the wire; between the plates, no physical current flows (it is a vacuum gap). Now apply Ampère's law (1) with an Amperian loop around the wire, and compute I_\text{enc} two ways:

Surface 1 — a flat disc capped on the loop, cutting through the wire. The disc is pierced by the wire, so I_\text{enc} = I. Ampère's law says \oint\vec{B}\cdot d\vec{\ell} = \mu_0 I.

Surface 2 — a bag-shaped surface capped on the same loop, but bulging out to pass between the capacitor plates. The bag does not cross the wire (it passes between the plates instead). So I_\text{enc} = 0. Ampère's law says \oint\vec{B}\cdot d\vec{\ell} = 0.

The same line integral cannot equal both \mu_0 I and zero. Ampère's law is inconsistent for this (time-varying) current. The problem: the capacitor's plates accumulate charge, so the electric field between them changes in time, but Ampère (1) has no term for this.

Maxwell's displacement-current fix

James Clerk Maxwell in 1861 resolved this by adding a term. The full statement becomes

\boxed{\;\oint\vec{B}\cdot d\vec{\ell} \;=\; \mu_0\,I_\text{enc} + \mu_0\,\varepsilon_0\,\frac{d\Phi_E}{dt}\;}

where \Phi_E = \int\vec{E}\cdot d\vec{A} is the electric flux through the capping surface. The new term \mu_0\varepsilon_0\,d\Phi_E/dt has the dimensions of \mu_0\timescurrent; Maxwell called the effective "current" I_d = \varepsilon_0\,d\Phi_E/dt the displacement current.

For the capacitor: on surface 1, \Phi_E = 0 (the disc is outside the plates, in the wire's region, where \vec{E} between the plates is inaccessible), d\Phi_E/dt = 0, and I_d = 0. Total: I_\text{enc} + I_d = I + 0 = I. \oint\vec{B}\cdot d\vec{\ell} = \mu_0 I. On surface 2, \Phi_E grows as the capacitor charges, and I_d works out to exactly I (because Q on the plates increases at rate I, and \Phi_E = Q/\varepsilon_0, so d\Phi_E/dt = I/\varepsilon_0, giving I_d = \varepsilon_0\cdot I/\varepsilon_0 = I). Total: I_\text{enc} + I_d = 0 + I = I. Same answer! The law is now consistent.

The displacement current is a real physical effect, not a bookkeeping trick

The term "displacement current" is slightly misleading — it is not a current of anything moving. It is a changing electric field acting like a current for the purpose of sourcing \vec{B}. But it has the same physical effect: it produces magnetic field. And that magnetic field is measurable, predictable, and real.

The deeper insight is this: a changing electric field creates a magnetic field, just as a current does. Symmetrically, Faraday's law already tells you that a changing magnetic field creates an electric field. Put these together:

  • Changing \vec{E} creates \vec{B} (Maxwell-Ampère).
  • Changing \vec{B} creates \vec{E} (Faraday).

This loop — a changing \vec{E} makes a \vec{B} that is also changing, which makes an \vec{E} that is also changing, which makes a \vec{B}... — admits self-consistent propagating solutions. Maxwell solved the equations and found: electric and magnetic fields can propagate together through empty space, at a speed c = 1/\sqrt{\mu_0\varepsilon_0} ≈ 3×10⁸ m/s. That is the speed of light. Light is an electromagnetic wave, and the displacement current is what makes it possible.

The modification from Ampère (1820) to Maxwell-Ampère (1861) took forty years but unified electricity, magnetism, and optics into a single framework. Every electromagnetic wave you use — from the 2.4 GHz WiFi router in your home to the visible light from the bulb in your ceiling to the gamma rays from the spent fuel in a BARC research reactor — is a solution to Maxwell's equations, and their existence depends on Maxwell's displacement current term.

The differential form and Stokes' theorem

Applying Stokes' theorem \oint\vec{B}\cdot d\vec{\ell} = \int(\nabla\times\vec{B})\cdot d\vec{A} to the integrated form, you get the local version of the Maxwell-Ampère law:

\nabla\times\vec{B} \;=\; \mu_0\vec{J} + \mu_0\varepsilon_0\,\frac{\partial\vec{E}}{\partial t}.

This is one of the four Maxwell equations. Together with Gauss's law for \vec{E}, Gauss's law for \vec{B} (\nabla\cdot\vec{B} = 0), and Faraday's law (\nabla\times\vec{E} = -\partial\vec{B}/\partial t), it forms the complete local description of classical electromagnetism — everything from transformers in the Indian power grid to the radiation from a Chandra-orbit X-ray telescope is governed by these four lines.

An Indian historical note

The differential form of Ampère's law in modern notation, with the displacement current included, is what Satyendra Nath Bose learned as a student at Presidency College, Calcutta, before he wrote the 1924 paper on photon statistics that became Bose-Einstein statistics. The equations he studied — the ones you are studying now — are the same ones that gave him the intuition to quantise light's degrees of freedom. Good physics education in India has always been this: master the classical equations deeply enough that the modern extensions feel like natural next steps, not new difficulties.

Where this leads next