SolenoidJEE Main

Uniform field inside a coil — B = μ₀nI — interactive Physics simulation for IIT-JEE.

Concept

A long solenoid concentrates its field inside, where it is remarkably uniform and axial: $B = \mu_0 n I$ . Outside, the field is nearly zero — the return flux spreads over all space. The geometry makes it the workhorse of electromagnets, inductors, relays and MRI bores.

Key formula

B_{inside} = \mu_0 n I, \qquad B_{end} = \tfrac{1}{2}\mu_0 n I, \qquad L = \mu_0 n^2 A\ell, \qquad U = \tfrac12 LI^2

Derivation

Ampère's law on a rectangular loop straddling the winding: only the inside segment (length ℓ) contributes, enclosing $n\ell$ turns:

$ $B\ell = \mu_0 (n\ell) I \Rightarrow B = \mu_0 n I$ $

Independent of radius and position (deep inside). At an end, symmetry halves it. Flux linkage $N\Phi = (n\ell)(BA)$ gives $L = \mu_0n^2A\ell$ .

Scenarios to explore

Solenoid — B = μ₀nI — uniform field, inductance & stored energy.

Real-world applications

Electromagnets, relays, solenoid valves & door locks.
Inductors in power supplies and filters.
MRI machines — superconducting solenoids at several tesla.

JEE exam tips

End field = half the centre field — instant answer via superposition of two half-solenoids.
Energy density u = B²/2μ₀ works anywhere, not just in solenoids.
Toroid: same formula with n = N/2πr — Ampère's law again.

Common mistakes

Using total turns N instead of turns per metre n in B = μ₀nI.
Thinking B depends on the solenoid's radius — it doesn't (long solenoid).
Doubling the current AND expecting quadrupled energy from B alone — U ∝ I², via ½LI².

Exam traps to avoid

Stretching a solenoid (same turns) reduces n, hence B drops.
An iron core multiplies B and L by μᵣ but saturates at high I in reality.