On this page
Inductance feels slippery because nothing happens while the current is steady. All the action is in the change. Once you treat an inductor as electrical inertia, most questions become simple: what does it do at the instant you close the switch, what does it do after a long time, and how much energy is stored? This post works through each with numbers.
What do the syllabi cover?
The JEE (Main) 2026 syllabus lists self and mutual inductance in its electromagnetic induction unit. The JEE (Advanced) 2026 brochure lists self and mutual inductance and RC, LR, LC and LCR circuits with d.c. and a.c. sources, so the exponential growth of current in an LR circuit is explicitly in scope there. NCERT Class 12 Chapter 6, section 6.7 derives the inductance of solenoids and the energy ½LI², and describes L as "electrical inertia" that "opposes growth and decay of current". Check the bulletin for your exam year.
How do you find self-inductance?
For a long solenoid of N turns, length l and cross-section A, the field inside is μ0(N/l)I and the flux linkage is N times BA, so L = μ0N²A/l.
Worked example. N = 500, l = 0.25 m, A = 2.0 × 10⁻⁴ m².
N²A = 250 000 × 2.0 × 10⁻⁴ = 50.
L = 4π × 10⁻⁷ × 50/0.25 ≈ 2.5 × 10⁻⁴ H, or 0.25 mH.
Doubling the turns quadruples L, because N appears once in the field and once in the linkage.
How much energy does an inductor store?
Building up a current I against the back emf takes work, stored in the magnetic field: U = ½LI². NCERT points out the parallel with ½mv², with L playing the part of mass.
Example. L = 0.20 H carries 3.0 A. U = ½ × 0.20 × 9.0 = 0.90 J.
What is mutual inductance?
When current in coil 1 changes, the flux through coil 2 changes, inducing an emf ε₂ = −M dI₁/dt.
NCERT exercise 6.8. Two coils have M = 1.5 H. The current in one goes from 0 to 20 A in 0.5 s.
Change in flux linkage with the other coil = MΔI = 1.5 × 20 = 30 Wb.
Average induced emf = 30/0.5 = 60 V.
How does current grow in an LR circuit?
Connect L = 0.20 H and R = 10 Ω in series with a 12 V battery and close the switch.
At t = 0: the current is zero, because it cannot jump. The whole 12 V appears across the inductor, so dI/dt = E/L = 60 A/s. The inductor acts like a break in the circuit.
After a long time: the current is steady, so the inductor has no emf. It acts like a plain wire. I = E/R = 1.2 A.
In between: I = (E/R)(1 − e^(−t/τ)) with time constant τ = L/R = 0.020 s.
At t = τ = 20 ms, I = 1.2 × (1 − e⁻¹) ≈ 1.2 × 0.632 ≈ 0.76 A.
Half the final current is reached at t = τ ln 2 ≈ 13.9 ms.
Final stored energy: ½ × 0.20 × 1.2² = 0.144 J.
If you then remove the battery and leave L and R in a closed loop, the current decays as I = I₀e^(−t/τ) with the same τ. Our post on RC circuits and charging uses the same exponential shape, and differential equations for JEE shows where it comes from.
Why do students mix up these ideas?
Electromagnetic induction is a known weak spot. Zuza and colleagues (2014) note that physics education research has shown its basic concepts are barely understood by students, and that teaching often moves quickly to rote problem solving. Earlier work by Bagno and Eylon (1997) found that students who finished a traditional high-school electromagnetism course had gaps in how their knowledge was organised, in conceptual understanding and in applying central relationships to problems. Students who solved problems, reflected on the concepts and built a concept map did better on recall, conceptual questions and problem solving.
For inductors, that means linking each formula to the one idea behind it: an inductor resists change in current.
Common mistakes
τ = LR instead of L/R. Check units: henry per ohm is seconds.
Current jumps at t = 0. It cannot. The voltage across the inductor can.
Energy from voltage. U = ½LI² uses current. There is no ½LV².
N instead of N² in L. Inductance goes as the square of the turns.
Back emf sign. The induced emf opposes the change, not the current.
A practice routine
Find L for three solenoids, then predict what happens to L if you double N, l or A.
Do four LR problems: at t = 0, at t = τ, at half the final current, and after a long time.
Do two mutual inductance problems and one where you find the energy stored at steady state.
If you are not sure where to start with this chapter, Fermi's AI tutor can work out what you already know and build a plan for the session, so you begin at the right step.
Tonight, sketch the two snapshots, t = 0 and t → ∞, for the 0.20 H and 10 Ω circuit without looking.
