Solving Energy Stored in Inductor Numericals (PYQ)
Physics · Electromagnetic Induction · NEET
Every energy-stored-in-inductor numerical uses one formula: U = ½LI² (energy in joules, L in henry, I in ampere). If they give you the energy and ask for L, just rearrange to L = 2U/I²; if they give L and I, plug straight in. Memory hook: "Half-L-I-squared" — the current is ALWAYS squared, just like ½mv² for kinetic energy.
Stored energy U = ½LI² grows as the square of current: doubling I gives 4× the energy. This is the parabola NTA exploits in trap options.
Your doubts, answered
Which formula do I use for energy stored in an inductor?
Only one: U = ½LI². Here U is the magnetic energy in joules (J), L is the inductance in henry (H), and I is the current in ampere (A). It comes from the work done to build up the current: W = ∫₀ᴵ L·I dI = ½LI². If a question gives you any two of U, L, I, you can find the third.
They gave me energy and current and asked for inductance. How?
Rearrange the same formula. From U = ½LI², solve for L: L = 2U / I². Example: U = 25 mJ, I = 60 mA gives L = (2 × 25×10⁻³) / (60×10⁻³)² = (50×10⁻³) / (3.6×10⁻³) ≈ 13.89 H. Notice the current in the denominator is squared before you divide.
How do I handle the mJ, µH, mA units without slipping?
Convert everything to SI base units FIRST, then plug in. milli (m) = ×10⁻³, micro (µ) = ×10⁻⁶. So 25 mJ = 25×10⁻³ J, 60 mA = 60×10⁻³ A, 4 µH = 4×10⁻⁶ H. Do the arithmetic in powers of ten, then convert the final answer back if the options are in mJ or µJ.
Why is the current squared and not the inductance?
Because energy builds up gradually as current rises from 0 to I. The average current during build-up is I/2, and the work is (average current) × (flux linkage), which brings in one factor of I from the current and one from the induced back-emf — giving I². L appears only to the first power. So doubling the current makes the stored energy four times bigger, not two times.
Is 'magnetic potential energy' the same as 'energy stored in the inductor'?
Yes. NEET questions phrase it as 'magnetic potential energy', 'magnetic energy stored', or 'energy stored in the inductor' — all three mean U = ½LI². Don't let the wording confuse you; the formula never changes.
⚠️ The NEET trap ✗ Forgetting to square the current, e.g. computing L = 2U/I = (50×10⁻³)/(60×10⁻³) ≈ 0.83 H, or matching the un-squared distractor. ✓ Square the current in the denominator: L = 2U/I² = (50×10⁻³)/(3.6×10⁻³) ≈ 13.89 H. Also remember to square the 10⁻³ (mA) into 10⁻⁶ (=×10⁻³ again after ÷ shift), which is exactly why the 0.138 H distractor exists. 🧠 The squaring trap: NTA hides the error inside the units.
Real NEET questions
NEET 2018
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance:
A · 1.389 H
B · 138.88 H
C · 0.138 H
D · 13.89 H ✓
Solution: Use U = ½LI², so L = 2U/I². Convert units: U = 25 mJ = 25×10⁻³ J, I = 60 mA = 60×10⁻³ A. Then I² = (60×10⁻³)² = 3.6×10⁻³ A². So L = (2 × 25×10⁻³) / (3.6×10⁻³) = (50×10⁻³)/(3.6×10⁻³) ≈ 13.89 H. Answer: D. (The 0.138 H distractor comes from mishandling the power of ten.)
NEET 2023
The magnetic energy stored in an inductor of inductance 4 µH carrying a current of 2 A is:
A · 4 J
B · 4 mJ
C · 8 mJ
D · 8 µJ ✓
Solution: Direct plug-in: U = ½LI². Convert L = 4 µH = 4×10⁻⁶ H, I = 2 A. Then U = ½ × (4×10⁻⁶) × (2)² = ½ × 4×10⁻⁶ × 4 = 8×10⁻⁶ J = 8 µJ. Answer: D. Key step: square the current (2² = 4) before multiplying.
Solved Electromagnetic Induction NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
