Ampere's Circuital Law

Q: What is Ampere's circuital law?

Ampere's law states that the line integral of B around any closed loop equals μ₀ times the net current enclosed: ∮B·dℓ = μ₀ I_enc. It is a fundamental relation between magnetic field and current, valid always, and useful for calculating B when the current distribution is highly symmetric.

Q: When can Ampere's law be used to calculate the magnetic field?

Only when an Amperian loop exists on which B has constant magnitude and is either parallel or perpendicular to dℓ. This requires strong symmetry: straight wires, solenoids, toroids, current sheets, cylinders. Otherwise, use Biot-Savart.

Q: What is the magnetic field inside a long solenoid?

Inside an ideal long solenoid with n turns per unit length carrying current I, B = μ₀nI, uniform and along the axis. Outside it is nearly zero. At either end (on axis), B drops to μ₀nI/2.

Q: Why is the magnetic field zero outside a toroid?

An Amperian loop outside the toroid encloses zero net current, and by symmetry B must be uniform along the loop, so B = 0 outside the toroid.

Q: What is the magnetic field inside and outside a solid current-carrying cylinder?

For a solid cylinder of radius R with uniform current density: inside (r R), B = μ₀I/(2πr), falling as 1/r. Maximum at r=R.

Q: How is Ampere's law analogous to Gauss's law?

Gauss's law relates electric flux through a closed surface to enclosed charge; Ampere's law relates circulation of B around a closed loop to enclosed current. Both are always true and both are useful for direct field calculation only under symmetry.

Q: Why is the field of an infinite current sheet independent of distance?

By symmetry, B is parallel to the sheet and constant on either side. An Amperian rectangle of length L straddling the sheet gives 2BL = μ₀KL, so B = μ₀K/2, with no dependence on distance.

Q: Is Ampere's law valid for time-varying currents?

In its original form, only for steady currents. For time-varying currents, Maxwell added a displacement-current term: ∮B·dℓ = μ₀(I + ε₀ dΦ_E/dt), known as the Ampere-Maxwell law.

PhysicsMagnetic Effects of Current and MagnetismFor JEE aspirants

Ampere's circuital law states that the line integral of $B$ around any closed loop equals $μ_{0}$ times the net current threading the loop: $\oint B \cdot d ℓ = μ_{0} I_{enc}$ . It is the magnetic analogue of Gauss's law in electrostatics and, when the current distribution has enough symmetry, gives the field far faster than Biot-Savart integration. For a JEE Main and NEET student, the four textbook applications are: the infinite straight wire, the long solenoid ( $B = μ_{0} n I$ ), the toroid, and current-carrying cylinders (solid and hollow). Ampere's law only calculates $B$ when a symmetric Amperian loop exists on which $B$ is constant and either parallel or perpendicular to $d ℓ$ ; otherwise, use Biot-Savart.

Key Formulas - Quick Reference

Ampere's law: $\oint B \cdot d ℓ = μ_{0} I_{enc}$
Infinite straight wire: $B = \frac{μ _{0} I}{2 π r}$
Long (ideal) solenoid, inside: $B = μ_{0} n I$ where $n = N / L$ = turns per unit length
Solenoid at either end: $B = \frac{1}{2} μ_{0} n I$
Toroid: $B = \frac{μ _{0} N I}{2 π r}$ inside, $B = 0$ outside
Solid cylinder of radius $R$ (uniform $J$ ): outside $r > R$ , $B = \frac{μ _{0} I}{2 π r}$ ; inside $r < R$ , $B = \frac{μ _{0} I r}{2 π R ^{2}}$
Hollow cylinder (surface current): $B = 0$ inside, $B = \frac{μ _{0} I}{2 π r}$ outside
Infinite current sheet, surface current density $K$ : $B = \frac{μ _{0} K}{2}$ on either side

Colours in figures: magnetic field

B

current

I

Amperian loop

1Statement of Ampere's Law

For any closed curve (Amperian loop) drawn in space, the line integral of the magnetic field along the curve equals $μ_{0}$ times the algebraic sum of currents piercing the surface bounded by the curve:

\oint_{C} B \cdot d ℓ = μ_{0} I_{enc}

Sign convention: curl the right-hand fingers along $d ℓ$ ; the thumb defines the positive normal direction. Currents flowing in the direction of this normal count as positive; currents flowing opposite count as negative.

Figure 1An Amperian loop traversed anticlockwise, so currents out of the page count as positive.

I_{1}

and

I_{3}

(out) are positive,

I_{2}

(into) is negative, and

I_{4}

lies outside and is not counted:

I_{enc} = I_{1} - I_{2} + I_{3}

I_{4}

still contributes to

B

at points on the loop.

Important:

\oint B \cdot d ℓ = 0

does not mean

B = 0

everywhere along the loop - it just means the net enclosed current is zero. And

B

at a point may have contributions from currents outside the loop; Ampere's law only uses the enclosed current for the integral, not for pointwise field.

2Application: Infinite Straight Wire

Draw a circular Amperian loop of radius $r$ coaxial with the wire. By symmetry, $B$ has constant magnitude on this circle and is tangent to it, so $B ∥ d ℓ$ :

$\oint B \cdot d ℓ = B (2 π r) = μ_{0} I ⟹ B = \frac{μ _{0} I}{2 π r}$

This is the same result as Biot-Savart's, obtained in one line rather than by integration.

Figure 2Circular Amperian loop of radius

r

around a long straight wire (current out of page). By symmetry

B

has the same magnitude everywhere on the loop and is parallel to

d ℓ

, so

\oint B \cdot d ℓ = B (2 π r) = μ_{0} I

3Application: Long (Ideal) Solenoid

An ideal solenoid is tightly wound with $n$ turns per unit length carrying current $I$ , in the limit that its length $≫$ its radius. The field inside is uniform and along the axis; outside it is essentially zero.

Figure 3Long solenoid in section: turns carry current out of the page along the top and into the page along the bottom. The rectangular loop

P QR S

has

P Q

(length

L

) inside, where

B

is uniform and parallel to the axis;

QR

and

S P

are perpendicular to

B

;

R S

lies outside, where

B \approx 0

Along $P Q$ of length $L$ , $B ∥ d ℓ$ , so $\int_{P}^{Q} B \cdot d ℓ = B L$ . Along $QR$ and $S P$ , $B ⊥ d ℓ$ , so contribution is 0. Along $R S$ (outside), $B \approx 0$ . Total enclosed current $= (n L) I$ . Applying Ampere's law:

$B L = μ_{0} (n L) I ⟹ B = μ_{0} n I$

At the end of a finite solenoid (still long enough for the ideal formula to hold near the middle), the field drops to

B = \frac{1}{2} μ_{0} n I

- exactly half the mid-point value. The mid-point value can also be derived from the axial-loop formula by integrating over the solenoid's turns.

Figure 4Field along the axis of a finite solenoid (length eight times its radius, computed exactly).

B

is almost exactly

μ_{0} n I

over the middle and falls to about

\frac{1}{2} μ_{0} n I

at each end face.

4Application: ToroidBeyond syllabus

A toroid is a solenoid bent into a doughnut shape with $N$ turns wound around a circular ring of mean radius $R$ . Draw a circular Amperian loop of radius $r$ through the interior of the coil:

Figure 5Toroid seen from above. Turns cross the inner rim out of the page and the outer rim into the page. Loop 1 (in the hole) encloses no current, so

B = 0

. Loop 2 (inside the core) encloses

N I

, so

B = \frac{μ _{0} N I}{2 π r}

. Loop 3 (outside) encloses

N I

out and

N I

in, so

B = 0

By symmetry, $B$ is constant along the loop and tangent to it. Enclosed current $= N I$ :

$B (2 π r) = μ_{0} N I ⟹ B = \frac{μ _{0} N I}{2 π r}$

Inside the coil (through the turns): $B = \frac{μ _{0} N I}{2 π r}$
Outside the toroid (Amperian loop encloses no net current since equal currents go in and out at every cross-section, or no current at all): $B = 0$
In the limit $r \approx R$ (ideal toroid), $B = μ_{0} n I$ with $n = N / (2 π R)$ , recovering the solenoid result.

5Application: Infinite Current-Carrying Sheet

An infinite plane sheet carries a surface current density $K$ (amperes per metre width). By symmetry $B$ is parallel to the sheet, perpendicular to $K$ , and opposite on the two sides. Draw a rectangular Amperian loop of length $L$ straddling the sheet:

$\oint B \cdot d ℓ = 2 B L = μ_{0} (K L) ⟹ B = \frac{μ _{0} K}{2}$

Note this is independent of distance from the sheet - just like the electric field of an infinite charged plane.

Figure 6Infinite sheet (edge-on) carrying surface current density

K

out of the page.

B

is parallel to the sheet and opposite on the two sides. For the rectangle, the two long sides each give

B L

and the short sides give zero, so

2 B L = μ_{0} K L

6Current-Carrying Cylinders (JEE Advanced favourite)

6.1Solid cylinder, uniform current density

A solid cylinder of radius $R$ carries total current $I$ uniformly distributed. Current density $J = I / (π R^{2})$ .

Outside ( $r > R$ ): loop encloses full current $I$ , so $B = \frac{μ _{0} I}{2 π r}$ (same as a thin wire).
Inside ( $r < R$ ): loop of radius $r$ encloses $I_{enc} = J π r^{2} = I r^{2} / R^{2}$ , so $B = \frac{μ _{0} I r}{2 π R ^{2}}$ . This grows linearly with $r$ from 0 at the axis to maximum $\frac{μ _{0} I}{2 π R}$ at the surface.

Figure 7Solid cylinder with uniform current density (current out of page). (a) Loop of radius

r < R

encloses only

I r^{2} / R^{2}

; loop of radius

r > R

encloses all of

I

. (b)

B

rises linearly inside, peaks at

\frac{μ _{0} I}{2 π R}

on the surface, and falls as

1/ r

outside.

6.2Hollow cylinder (thin shell of radius $R$ )

A thin cylindrical shell carrying current $I$ on its surface:

Inside ( $r < R$ ): loop encloses no current, so $B = 0$ .
Outside ( $r > R$ ): $B = \frac{μ _{0} I}{2 π r}$ .

Figure 8

B

against

r

for a thin hollow cylinder (solid line) compared with a solid cylinder of the same radius and current (dashed). Inside the shell

B = 0

; the field jumps to

\frac{μ _{0} I}{2 π R}

at the surface; outside, both give

B = \frac{μ _{0} I}{2 π r}

Solved Example 1

A coaxial cable has an inner solid wire of radius

a

carrying current

I_{0}

up, and an outer thin cylindrical shell of radius

c

carrying current

I_{0}

down. Find

B

at (i)

r < a

, (ii)

a < r < c

, (iii)

r > c

Figure 9Coaxial cable in section: inner solid conductor of radius

a

carries

I_{0}

out of the page; thin outer shell of radius

c

carries

I_{0}

into the page. Concentric Amperian loops in the three regions give the three results of the example.

Solution:

By symmetry, Amperian loops are circles concentric with the cable.

(i) $r < a$ : $I_{enc} = I_{0} (r / a)^{2}$ , so $B = \frac{μ _{0} I _{0} r}{2 π a ^{2}}$ .

(ii) $a < r < c$ : $I_{enc} = I_{0}$ , so $B = \frac{μ _{0} I _{0}}{2 π r}$ .

(iii) $r > c$ : $I_{enc} = I_{0} - I_{0} = 0$ , so $B = 0$ . The outer shell shields all external field - the reason coaxial cables have low EMI.

Solved Example 2

A solenoid of length

0.4 m

1000

turns, carries

5 mA

. Find the field at the middle and at the end.

Solution:

Turns per unit length: $n = 1000/0.4 = 2500 m^{- 1}$ .

Middle (ideal formula, since length $≫$ diameter for typical solenoid): $B = μ_{0} n I = (4 π \times 1 0^{- 7}) (2500) (5 \times 1 0^{- 3}) = 5 π \times 1 0^{- 6} \approx 1.57 \times 1 0^{- 5} T$ .

End: $B = \frac{1}{2} μ_{0} n I \approx 7.85 \times 1 0^{- 6} T$ .

Solved Example 3

Three identical long solenoids P, Q, R are connected: current

I

enters P, splits equally between Q and R (which are in parallel). The field at the centre of P is

2.0 T

. Find the field at the centre of Q.

Solution:

Since $B = μ_{0} n I$ and $n$ is the same for identical solenoids, the field scales linearly with the current through that solenoid. Currents through Q and R are each $I /2$ , so $B_{Q} = B_{P} /2 = 1.0 T$ .

7When to Use Ampere's Law vs Biot-Savart

Use Ampere's law when	Use Biot-Savart when
High symmetry (cylindrical, planar, toroidal)	Arbitrary geometry - single loop, arc, finite wire
You can pick a loop where $B$ is constant and $∥$ or $⊥$ to $d ℓ$	Field varies along any conceivable loop
Straight wire (long), solenoid, toroid, sheet, cylinder	Loop centre, loop axis, finite wire, arc, off-axis

Common Mistakes to Avoid

Watch out

Applying Ampere's law to any current geometry: the law is always true but only useful if symmetry lets you pull $B$ out of the line integral. Without symmetry, you get one equation with two unknowns ( $B$ direction and magnitude).
Confusing $\oint B \cdot d ℓ = 0$ with $B = 0$ : a loop with no enclosed current has zero line integral, but $B$ at points on the loop can be non-zero (produced by currents outside).
Wrong sign of enclosed current: always fix $d ℓ$ direction first, then apply right-hand rule to define the positive-current direction.
Using $B = μ_{0} n I$ for a short solenoid: valid only when length $≫$ diameter, deep inside. Near the ends or for short coils, use the general axial formula from Biot-Savart.
Confusing $n$ (turns per unit length) with $N$ (total turns) in the solenoid formula.
Assuming toroid field is uniform: $B \propto 1/ r$ inside the toroid, so field is stronger near the inner edge than the outer edge. Only in the thin-toroid limit is it nearly uniform.

Frequently Asked Questions

Q1. What is Ampere's circuital law?

Ampere's law states that the line integral of $B$ around any closed loop equals $μ_{0}$ times the net current enclosed: $\oint B \cdot d ℓ = μ_{0} I_{enc}$ . It is a fundamental relation between magnetic field and current, valid always, and useful for calculating $B$ when the current distribution is highly symmetric.

Q2. When can Ampere's law be used to calculate the magnetic field?

Ampere's law calculates $B$ only when you can find an Amperian loop on which $B$ has constant magnitude and is either parallel or perpendicular to $d ℓ$ . This requires strong symmetry: straight wires, solenoids, toroids, current sheets, cylinders. For arbitrary shapes, Biot-Savart is the right tool.

Q3. What is the magnetic field inside a long solenoid?

Inside an ideal long solenoid with $n$ turns per unit length carrying current $I$ , the field is uniform and along the axis: $B = μ_{0} n I$ . Outside, it is essentially zero. At either end (still on the axis), the field drops to $B = μ_{0} n I /2$ .

Q4. Why is the magnetic field zero outside a toroid?

An Amperian loop outside the toroid either (a) encloses no wire at all if it lies beyond the coil, or (b) encloses each turn twice, once going in and once going out, giving zero net enclosed current if you use the internal loop. Either way, symmetry plus zero net enclosed current gives $B = 0$ outside.

Q5. What is the magnetic field inside and outside a solid current-carrying cylinder?

For a solid cylinder of radius $R$ with uniform current density: inside ( $r < R$ ), $B = \frac{μ _{0} I r}{2 π R ^{2}}$ (grows linearly from zero at the axis); outside ( $r > R$ ), $B = \frac{μ _{0} I}{2 π r}$ (falls as $1/ r$ , same as a thin wire). The field is maximum at $r = R$ .

Q6. Does $\oint B \cdot d ℓ = 0$ mean $B = 0$ everywhere on the loop?

No. It means only that the net current enclosed by the loop is zero. $B$ at individual points on the loop can be non-zero, produced by currents outside the loop. The line integral cancels because contributions of opposite sign balance.

Q7. How is Ampere's law analogous to Gauss's law?

Gauss's law in electrostatics relates the electric flux through a closed surface to the enclosed charge: $\oint E \cdot d A = Q / ε_{0}$ . Ampere's law relates the circulation of $B$ around a closed loop to the enclosed current: $\oint B \cdot d ℓ = μ_{0} I$ . Both are always true; both are useful for direct field calculation only under symmetry.

Q8. Why is the field of an infinite current sheet independent of distance?

By symmetry, $B$ from an infinite sheet is parallel to the sheet, perpendicular to the current direction, and constant on either side. An Amperian rectangle of length $L$ straddling the sheet gives $2 B L = μ_{0} K L$ , so $B = μ_{0} K /2$ - no $r$ dependence. This mirrors the field of an infinite plane of charge in electrostatics.

Q9. Is Ampere's law valid for time-varying currents?

In the original form $\oint B \cdot d ℓ = μ_{0} I_{enc}$ , it holds only for steady (DC) currents. For time-varying currents (like charging a capacitor), Maxwell added a displacement-current term giving the Ampere-Maxwell law: $\oint B \cdot d ℓ = μ_{0} (I + ε_{0} d Φ_{E} / d t)$ . This is a Class 12 EMI-onwards topic.

Previous year questions on Ampere's Circuital Law

8 questions from past papers, each with a step-by-step solution.

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1Statement of Ampere's Law

2Application: Infinite Straight Wire

3Application: Long (Ideal) Solenoid

4Application: ToroidBeyond syllabus

5Application: Infinite Current-Carrying Sheet

6Current-Carrying Cylinders (JEE Advanced favourite)

6.1Solid cylinder, uniform current density

6.2Hollow cylinder (thin shell of radius R)

7When to Use Ampere's Law vs Biot-Savart

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is Ampere's circuital law?

Q2. When can Ampere's law be used to calculate the magnetic field?

Q3. What is the magnetic field inside a long solenoid?

Q4. Why is the magnetic field zero outside a toroid?

Q5. What is the magnetic field inside and outside a solid current-carrying cylinder?

Q6. Does ∮B⋅dℓ=0 mean B=0 everywhere on the loop?

Q7. How is Ampere's law analogous to Gauss's law?

Q8. Why is the field of an infinite current sheet independent of distance?

Q9. Is Ampere's law valid for time-varying currents?

Previous year questions on Ampere's Circuital Law

Ready to master Magnetic Effects of Current and Magnetism?

6.2Hollow cylinder (thin shell of radius $R$ )

Q6. Does $\oint B \cdot d ℓ = 0$ mean $B = 0$ everywhere on the loop?