JEE Mains Physics · Moving Charges and Magnetism

Ampère's Law: Thick Wires and Solenoids

Round any closed loop, the sum of B·dl is μ₀ times the current the loop encloses; with a symmetric shape this gives the field inside and outside a thick wire, a cable and a solenoid in one line.

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Why this matters

Twenty-three PYQs, nineteen of them multiple choice, and one from 2026. Thirteen apply Ampère's law to thick conductors: seven are about a solid wire's field inside and outside it, often as a graph, two are coaxial cables, one is a hollow tube, one a ring of rotating charges, and two test the law itself in statements or a match list. Ten are solenoids: four use B = μ₀nI directly, two ask for the magnetic intensity H = nI, two add an iron core, one has an electron circling inside, and one connects four solenoids in a network.

Concept 1 of 2: Ampère's law for a solid wire, a hollow tube and a coaxial cable

Ampère's law says the field round a closed path is set only by the current passing through that path. For a long round conductor, take a circle centred on the axis: B is the same all round it, so B times 2πr equals μ₀ times the current inside the circle. Inside a solid wire, a smaller circle encloses less current, so the field grows from zero at the axis to a maximum at the surface, then falls as 1/r outside.

Definition

Ampère's circuital law: $\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enc}}$ , for steady currents. It follows from the Biot–Savart law; it is not an independent result.
Solid wire of radius a, current I spread uniformly: inside, $I_{\text{enc}} = I\dfrac{r^{2}}{a^{2}}$ , so $B = \dfrac{\mu_0 I r}{2\pi a^{2}}$ (proportional to r, zero on the axis).
Outside the wire: $B = \dfrac{\mu_0 I}{2\pi r}$ . The largest field is at the surface, $\dfrac{\mu_0 I}{2\pi a}$ . Outside, the field does not depend on the wire's thickness.
Hollow tube with the current on its surface: B = 0 inside, $\dfrac{\mu_0 I}{2\pi r}$ outside, a jump at the surface.
Coaxial cable with equal and opposite currents in the inner wire and the outer shell: an Amperian circle outside both encloses zero net current, so B = 0 outside the cable.
Charges going round a circle act as a current: current = total charge × revolutions per second. An Amperian loop counts it once if the current passes through the loop once, and zero if it passes in and back out.

Ampère's law and a solid wire

\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enc}} \qquad B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^{2}},\quad B_{\text{out}} = \frac{\mu_0 I}{2\pi r}

Worked example

A long straight wire of radius 2 mm carries 10 A spread uniformly over its cross-section. Find the field at 0.5 mm, at 2 mm and at 5 mm from the axis. (

\mu_0/2\pi = 2 \times 10^{-7}

T m/A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q9Moderate

Example 1 · Moving Charges and Magnetism · Ampere's Law: Thick Wires and Solenoids

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be

Inside a solid wire B grows with r

Inside a wire with uniform current, B = μ₀Ir/2πa², which rises from zero at the axis. Using μ₀I/2πr inside gives a field that blows up at the axis, the opposite shape.

The field outside a coaxial cable is zero

With equal and opposite currents in the inner wire and the shell, a circle outside both encloses no net current, so B = 0 there. Between the conductors only the inner current counts.

Only enclosed current counts

Current flowing just outside an Amperian loop changes B at points on the loop but not the line integral of B round it. The integral is μ₀ times the current that passes through the loop.

Drill 12 more on ampère's law for a solid wire, a hollow tube and a coaxial cable

Concept 2 of 2: Field inside a solenoid and a toroid

A long, tightly wound solenoid has a strong, uniform field inside along its axis and almost none outside. Ampère's law round a rectangle that runs inside and comes back outside gives B times the length equal to μ₀ times the current through all the turns in that length. So the field depends only on turns per metre and the current, not on the radius.

Definition

Long solenoid: $B = \mu_0 n I$ , where n is turns per METRE (n = N/L). Uniform inside, about half that value at the ends, close to zero outside.
Magnetic intensity: $H = nI$ , in A/m. It does not include $\mu_0$ .
With a core of relative permeability $\mu_r$ : $B = \mu_0\mu_r n I$ .
Toroid of N turns, at radius r inside its core: $B = \dfrac{\mu_0 N I}{2\pi r}$ ; zero outside.
Identical solenoids give fields in proportion to the current through each, so in a network split the current first.
Flux through one cross-section is BA; flux linkage of the whole winding is NBA. Read which one a question asks for.

Solenoid, intensity, core and toroid

B = \mu_0 n I \qquad H = nI \qquad B = \mu_0\mu_r n I \qquad B_{\text{toroid}} = \frac{\mu_0 N I}{2\pi r}

Worked example

A solenoid 50 cm long has 400 turns and carries 3 A. Find (a) n, (b) the field inside, (c) H, and (d) the field if an iron core of relative permeability 200 fills it. (

\mu_0 = 4\pi \times 10^{-7}

T m/A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q13Moderate

Example 2 · Moving Charges and Magnetism · Ampere's Law: Thick Wires and Solenoids

A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube

2\text{ }m

long and

0.2\text{ }m

in diameter. The magnetic intensity at the center of the solenoid when a current of

2\text{ }A

flows through it is?

Turns per centimetre must become turns per metre

In B = μ₀nI, n is per metre. A winding of 20 turns per cm is n = 2000 per metre; leaving it as 20 makes B a hundred times too small.

H has no μ₀ in it

Magnetic intensity inside a solenoid is H = nI, in A/m. B = μ₀nI is in tesla. A question that asks for magnetic intensity wants H.

Flux and flux linkage differ by N

The flux through one cross-section of a solenoid is BA. The flux linked with the whole winding is NBA. The two answers differ by the number of turns, and both appear among the options.

Drill 9 more on field inside a solenoid and a toroid

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Ampère's law for a solid wire, a hollow tube and a coaxial cable
Ampère's law and a solid wire
$\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enc}} \qquad B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^{2}},\quad B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$
Field inside a solenoid and a toroid
Solenoid, intensity, core and toroid
$B = \mu_0 n I \qquad H = nI \qquad B = \mu_0\mu_r n I \qquad B_{\text{toroid}} = \frac{\mu_0 N I}{2\pi r}$

Watch out for (6)

Inside a solid wire B grows with r
→ Ampère's law for a solid wire, a hollow tube and a coaxial cable
The field outside a coaxial cable is zero
→ Ampère's law for a solid wire, a hollow tube and a coaxial cable
Only enclosed current counts
→ Ampère's law for a solid wire, a hollow tube and a coaxial cable
Turns per centimetre must become turns per metre
→ Field inside a solenoid and a toroid
H has no μ₀ in it
→ Field inside a solenoid and a toroid
Flux and flux linkage differ by N
→ Field inside a solenoid and a toroid

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