Q.A solenoid of length 50cm, having 100 turns carries a current of 2.5A. Find the magnetic field (B),
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Magnetic Field of a Solenoid
Imagine you take a long wire and wind it into a tight, helical coil — that's a solenoid. If you've ever seen a spring, it looks just like that. Now pass a current through this coil. Each loop of wire produces its own tiny magnetic field, and when you pack many loops close together, those individual fields add up.
The key insight is superposition: the total field at any point is the vector sum of the fields from every turn. Inside the coil, the fields from adjacent turns reinforce each other. Outside, they tend to cancel. The result is striking — a long solenoid behaves almost like a bar magnet, with a uniform field inside and a very weak field outside.
The solenoid is the magnetic analogue of a parallel-plate capacitor: it creates a uniform field in a confined region.
The Precise Statement
For an ideal solenoid — infinitely long, tightly wound, with negligible spacing between turns — the magnetic field is:
- Inside: uniform, directed along the axis of the solenoid, with magnitude
B=μ0nI
where n=N/L is the number of turns per unit length, I is the current, and μ0=4π×10−7T⋅m/A is the permeability of free space.
- Outside: nearly zero (strictly zero for an ideal solenoid; for a real one, it's very small and falls off rapidly with distance).
The direction of the field inside follows the right-hand rule: curl your fingers in the direction of the current around the coil, and your thumb points along the field inside.
Binside=μ0nI
Why Is the Field Uniform Inside?
Consider a rectangular Amperian loop that runs along the axis inside the solenoid, goes out radially, runs parallel to the axis outside, and returns. Ampère's law says:
∮B⋅dl=μ0Ienc
The outside field is negligible, so only the inside segment contributes. The enclosed current is nLI for a loop of axial length L. This gives:
BL=μ0(nLI)⇒B=μ0nI …
Deep inside a long solenoid the field is the uniform value set by the turns per unit length and the current, while an end point effectively sees only half the winding, giving half that field. …
Using B=μ0nI for the interior of a long solenoid, and half that value at the end: Binterior≈6.28×10−4 T, Bend≈3.14×10−4 T.
Given: length L=50 cm =0.5 m, turns N=100, current I=2.5 A, μ0=4π×10−7 Wb A⁻¹m⁻¹.
Turns per unit length: n=LN=0.5100=200 turns/m.
(a) Interior (deep inside, on the axis) of the solenoid: the field is essentially uniform and given by:
B=μ0nI=(4π×10−7)(200)(2.5)
=4π×10−7×500=2000π×10−7≈6.28×10−4 T
…
- CBSE 2026Set A1 markMCQQ.The magnetic field lines inside a current-carrying solenoid are (A) circular (B) diverging (C) parabola (D) parallel and straight
›Reveal solutionSolution
A long current-carrying solenoid produces a uniform axial field inside, so its field lines are parallel and straight.
Each turn of a solenoid behaves like a small circular loop. When many turns are stacked closely, their fields add along the axis and largely cancel outside. The result inside a long solenoid is a nearly uniform magnetic field of magnitude
B=μ0nI …
- CBSE 2026Set ANNUAL1 markQ.Match the Column A with Column B and write the correct pair. Column A: Magnetic field inside solenoid. Column B:(i) μ₀nI,(ii) Volt × second,(iii) μ₀nI/2,(iv) Volt × meter,(v) Volt × meter⁻¹,(vi) Volt.
›Reveal solutionSolution
Interior field of a long solenoid B = μ₀nI, option (i).
Using Ampere's circuital law for a long solenoid, the magnetic field well inside (away from the ends) is uniform and directed along the axis:
B = μ₀ n I,
…
- CBSE 2026Set ANNUAL1 markQ.Match the Column A with Column B and write the correct pair. Column A: Magnetic field at any end of solenoid. Column B:(i) μ₀nI,(ii) Volt × second,(iii) μ₀nI/2,(iv) Volt × meter,(v) Volt × meter⁻¹,(vi) Volt.
›Reveal solutionSolution
At a solenoid's end the field is half the interior value: B_end = μ₀nI/2, option (iii).
For a long solenoid, the interior field is B = μ₀nI. At the very end of the solenoid, only 'half' of the contributing turns lie on one side, and the field on the axis drops to exactly half …
- CBSE 2026Set SEM31 markMCQQ.If the radius of an ideal solenoid be r and magnetic field intensity on its axis according to Ampere's circuital law be B, then(a) B ∝ r(b) B ∝ 1/r(c) B ∝ 1/r²(d) B is independent of r
›Reveal solutionSolution
Ampère's circuital law gives the axial field of a long ideal solenoid as B = μ₀nI, which contains no radius term, so B does not depend on r. Option (d).
Step 1 — apply Ampère's law to a rectangular loop, partly inside and partly outside a long ideal solenoid (a standard NCERT/CBSE Class 12 Physics derivation in Moving Charges and Magnetism).
…
- CBSE 2025Set D1 markMCQQ.The direction of magnetic field inside a current carrying solenoid is (A) circular (B) parallel to axis (C) perpendicular to the axis (D) random
›Reveal solutionSolution
The field inside a solenoid is uniform and parallel to its axis.
A solenoid is a long coil of many turns. The circular fields of the individual turns add up so that, well inside a long solenoid, the resultant magnetic field is nearly uniform, strong, and directed along the axis (its direction given by the right-hand rule). Outside …
- CBSE 2025Set ANNUAL1 markMCQQ.A uniform magnetic field is produced(a) from straight current carrying conductor(b) at the centre of a current carrying circular loop(c) at the axis of a current carrying circular loop(d) inside a current carrying solenoid
›Reveal solutionSolution
A long, tightly-wound solenoid is the standard example of a uniform magnetic field region: inside it (away from the ends), B is constant in both magnitude and direction.
Compare the field patterns:
- A straight current-carrying wire produces a field that varies as 1/r with distance and circles the wire (non-uniform).
- At the centre of a circular loop the field has a single fixed value only at that one point, not over a region. …
- CBSE 2022Set HE2171 markQ.Fill in the blank: The behavior of Solenoid is like as ______.
›Reveal solutionSolution
A current-carrying solenoid behaves like a bar magnet, with one end acting as a north pole and the other as a south pole.
When a steady current flows through a solenoid (a long, closely wound helical coil), the magnetic field lines emerge from one end and enter the other, exactly as they do for a bar magnet, and the field pattern outside the solenoid is indistinguishable from that of an equivalent bar magnet of the same magnet …
- CBSE 2020Set ANNUAL1 markQ.Give any one use of electromagnet
›Reveal solutionSolution
Electromagnets give a strong magnetic field only while current flows, so they are used where a switchable magnet is needed — e.g. lifting heavy iron loads in cranes.
Concept. When current passes through a solenoid wound on a soft-iron core, the core becomes strongly magnetised, producing an electromagnet. The magnetism can be switched on/off and its strength controlled by the current.
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