Ampere's Law, Solenoid and Toroid

• State and apply Ampere's circuital law to find B in symmetric systems
• Derive the uniform magnetic field inside a solenoid
• Derive the field inside a toroid and explain why it is zero outside
• Compare solenoids and toroids as field-producing devices
• Calculate n (turns per unit length) from given winding data
• Avoid common errors in choosing the Amperian loop

Level 1 — Foundations

Ampere's Circuital Law: The line integral of the magnetic field B around any closed loop (Amperian loop) equals μ₀ times the total current enclosed by that loop:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$

This is the magnetic analogue of Gauss's law. It is exact only for steady (DC) currents; Maxwell later added the displacement current term for time-varying fields.

Key rule: Only the current passing through the area bounded by the chosen loop contributes. Direction of current is judged by the right-hand rule relative to the chosen sense of traversal.

Solenoid: A long coil of wire wound in a helix. For an ideal solenoid (infinite length, closely wound):

• Inside: B = μ₀nI (uniform, parallel to axis)
• Outside: B = 0

where n = N/L = number of turns per unit length.

Toroid: A solenoid bent into a closed doughnut shape (torus):

• Inside the toroid (in the gap): B = μ₀nI = μ₀NI/(2πr)
• Outside (centre or exterior): B = 0

where r is the distance from the centre of the toroid to the Amperian loop.

Level 2 — JEE depth

Proof: Solenoid field using Ampere's law Choose a rectangular Amperian loop ABCD with AB (length L) inside the solenoid parallel to the axis, and CD outside. The field outside is zero; the transverse sides BC and DA contribute nothing (B ⊥ dl there). Therefore:

$\oint \vec{B} \cdot d\vec{l} = B \cdot L$

Enclosed current = (nL) turns × I each = nLI

$B \cdot L = \mu_0 n L I \implies \boxed{B = \mu_0 n I}$

This result is independent of position inside the solenoid — the field is perfectly uniform for an ideal solenoid.

B at an interior point of a toroid: Choose an Amperian circle of radius r (mean radius) inside the toroid. All N turns thread through this loop:

$B \cdot 2\pi r = \mu_0 N I \implies B = \frac{\mu_0 N I}{2\pi r}$

Unlike the solenoid, the field inside a toroid varies with r (inversely).

Why B = 0 outside the toroid:

• For a loop outside the toroid (r > outer radius): net enclosed current = N turns going in one direction through the loop — but the toroid's winding means the same N turns also come back through the loop in the other direction. Net enclosed current = 0, so B = 0.
• For a loop at the centre hole (r < inner radius): no turns enclosed, so B = 0.

Comparison: Solenoid vs Toroid

Finding B inside/outside a cylindrical conductor of radius R carrying I:

• r > R (outside): ∮B·dl = μ₀I → B = μ₀I/(2πr) (same as wire)
• r < R (inside, uniform current density): I_enc = I(r²/R²) → B = μ₀Ir/(2πR²)
• B is maximum at the surface (r = R) and zero at r = 0.

JEE traps:

• n is turns per unit length (N/L), not N alone. Always compute n from given data.
• Ampere's law gives the correct answer only if B is constant in magnitude and parallel/perpendicular to dl along the chosen loop — if symmetry is absent, use Biot-Savart.
• For a solenoid of finite length, the field at the ends is μ₀nI/2 (half of the interior value).
• The toroid formula B = μ₀NI/(2πr) uses mean radius for quick estimates but strictly r is the variable radius.

Solenoid — Field Inside

Given: N = 500 turns, L = 40 cm = 0.40 m, I = 2 A
Step 1: Find n (turns per unit length)
  n = N/L = 500 / 0.40 = 1250 turns/m

Step 2: Apply solenoid formula
  B = μ₀nI
  B = (4π × 10⁻⁷) × 1250 × 2

Step 3: Calculate
  B = 4π × 10⁻⁷ × 2500
  B = 10000π × 10⁻⁷
  B = π × 10⁻³
  B ≈ 3.14 × 10⁻³ T

Answer: B ≈ 3.14 mT, uniform inside the solenoid, parallel to axis.

Toroid — Field Inside

Given: N = 1000 turns, mean radius r = 20 cm = 0.20 m, I = 3 A
Step 1: Apply toroid formula
  B = μ₀NI / (2πr)

Step 2: Substitute
  B = (4π × 10⁻⁷ × 1000 × 3) / (2π × 0.20)

Step 3: Simplify
  Numerator: 4π × 10⁻⁷ × 3000 = 12000π × 10⁻⁷
  Denominator: 2π × 0.20 = 0.4π

  B = (12000π × 10⁻⁷) / (0.4π)
  B = 12000 × 10⁻⁷ / 0.4
  B = 30000 × 10⁻⁷
  B = 3.0 × 10⁻³ T

Answer: B = 3.0 mT inside the toroid (at mean radius 20 cm).
        B = 0 at the centre and outside the toroid.

• Q1: A solenoid has 800 turns, length 80 cm, and current 1.5 A. Find B inside.
• Q2: For a toroid with N = 600, mean radius 15 cm, I = 4 A — find B and compare to the same parameters in a straight solenoid.
• Q3: A thick cylindrical conductor of radius 5 mm carries 10 A uniformly. Find B at r = 2 mm and at r = 8 mm.
• Q4: Why does Ampere's law give B = 0 inside the hole of a toroid?
• Stretch: Q5: A solenoid of 2000 turns, 1 m long, 4 cm diameter, carries 5 A. Find the total flux through all turns (flux linkage NΦ) and verify units are Wb.

NCERT Chapter 4 link: Section 4.7 covers Ampere's circuital law; Sections 4.8–4.9 apply it to the solenoid and toroid with full diagrams. The finite solenoid field at the end is also discussed. All NCERT exercises in this chapter should be solved using Ampere's law first — then verify with Biot-Savart where feasible.

Exam connections: JEE tests: (1) choosing the correct Amperian loop for non-obvious geometries (coaxial cable), (2) comparing B inside/outside a conductor, (3) numerical problems needing n = N/L conversion, (4) the concept that B = 0 at the toroid centre as an MCQ trap.

Study strategy: For every problem, draw the cross-section, mark the current directions, draw the Amperian loop, and explicitly state I_enc before writing any formula. This three-step discipline eliminates 80% of common errors on Ampere's law questions.

• Use the Solenoid/Toroid live world: vary N, L, and I with sliders; observe how the internal field changes and confirm B = 0 outside the toroid by moving a virtual field probe.
• Home activity: Wind two sets of 20 turns each around a cardboard tube — one with equal spacing (solenoid), one bent into a ring (approximate toroid). Use a compass to probe inside/outside each and record observations.
• AI Mentor voice reflection: "Explain to a family member why a toroid is preferred over a solenoid in your mobile phone charger circuit."

• "Explain Ampere's law to a Class 10 student by comparing it to how you find the 'total crowd' inside a fence rather than counting every person individually."
• "What is one common mistake when calculating n for a solenoid, and how would you catch yourself making it?"
• Stretch: "How does the toroid's ability to contain the magnetic field completely connect to reducing electromagnetic interference (EMI) in electronics and the design of wireless chargers?"

• Complete the core practice + one extension activity (photo, table, short reflection, or mini-project) for base XP + topic badge.
• 5-7 day streak or family discussion note = multiplier + visible artifact in parent/principal dashboard.
• Best real-world application stories (anonymised) featured on class or national leaderboard.

• Build a solenoid switch: wind 100 turns of wire on a pen body, pass current from two AA batteries, and lift a small iron screw — measure the current and compute B.
• Direct link to Cyber Defenders (EMI shielding in circuit design) and Micro-Entrepreneurship (understanding inductive charging products).
• Coding extension: Simulate Ampere's law with a Python loop — calculate the enclosed current for various loop radii around a conductor and plot B vs r.

This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.

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Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.

See curriculum for cross-links and the full future-skills/robotics chapters.