Induced Electric Field from a Changing Magnetic Field
Physics · Electromagnetic Induction · NEET
When the magnetic field inside a region changes with time, it creates an electric field in the space around it, even with no wire and no charges present. This induced electric field forms closed loops (it is non-conservative), and its line integral equals the rate of change of magnetic flux: ∮ E · dl = -dΦ/dt. Memory hook: "changing B breeds curly E" — a shrinking or growing magnetic field always wraps a circular electric field around itself.
Left: a magnetic field growing inside a circle of radius r wraps a closed (non-conservative) induced electric field E around it — no wire needed. Right: emf of a loop uses only the flux area πr² where B exists; a loop entirely outside the field region encloses zero flux, so its emf is zero.
Your doubts, answered
Does a changing magnetic field really make an electric field even with no wire?
Yes. This is the deepest idea in Faraday's law. A wire and its induced current are just what we SEE. The real cause is that a time-varying magnetic field sets up an electric field in the empty space around it. If you put a conducting loop there, this E field pushes charges and you get a current. But the E field exists whether or not the wire is present. For NEET: emf can exist with no charges and no conductor — the changing B is enough.
Is this induced electric field the same as the field from a charge?
No, and this is a common trap. A field from a static charge is conservative — its field lines start on + charge and end on - charge, and ∮ E · dl = 0. The induced electric field from a changing B is non-conservative — its field lines are closed loops (no start, no end), and ∮ E · dl = -dΦ/dt is NOT zero. So the concept of potential (voltage) does not apply cleanly to an induced electric field.
How do I find the strength of E inside a region where B is changing?
Use ∮ E · dl = -dΦ/dt on a circular Amperian-style loop of radius x. By symmetry E is constant along the circle, so E × (2πx) = (dB/dt) × (area of flux enclosed). Inside a region of radius r (for x < r): E × 2πx = πx² × dB/dt, giving E = (x/2)(dB/dt). Outside (x > r) the enclosed flux is fixed at πr², so E = (r²)/(2x) × dB/dt — E falls off like 1/x.
If a loop is placed OUTSIDE the field region, why can the emf still be zero?
The emf around any loop equals -dΦ/dt where Φ is the flux THROUGH that loop. If the whole changing-B region lies outside the loop, the loop encloses zero flux, so dΦ/dt = 0 and emf = 0 — even though an induced E field does exist along the loop. The tangential E fields around the loop cancel out when you go all the way around. This is exactly the NEET 2016 two-loop question.
Why is there no minus sign confusion here?
The minus sign (Lenz's law) only tells the DIRECTION of the induced E field — it opposes the change in flux. For magnitude questions you drop the sign and just use |emf| = |dΦ/dt|. NEET numericals almost always ask for magnitude, so compute the value, then check direction separately only if asked.
⚠️ The NEET trap ✗ For loop 1 (radius R) enclosing a field region of radius r, students write emf = -(dB/dt)πR² using the loop's own area R². ✓ emf = -(dB/dt)πr². Only the area where B actually exists (radius r) carries flux; the empty ring between r and R has no B, so it adds no flux. Use the FLUX area, not the loop area. 🧠 Flux lives only where B lives — multiply dB/dt by the field's area πr², never by the loop's area πR².
Real NEET questions
NEET 2016 Phase 2
A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate dB/dt. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field. Then the e.m.f. generated is
A · Zero in loop 1 and zero in loop 2
B · -(dB/dt)πr² in loop 1 and -(dB/dt)πr² in loop 2
C · -(dB/dt)πR² in loop 1 and zero in loop 2
D · -(dB/dt)πr² in loop 1 and zero in loop 2 ✓
Solution: Step 1 — emf = -dΦ/dt, where Φ is the flux enclosed by each loop. Step 2 — Loop 1 (radius R > r) surrounds the whole field region, so it encloses flux Φ = B × πr² (the field exists only up to radius r). emf₁ = -dΦ/dt = -(dB/dt)πr². Note we use πr², NOT πR², because there is no B in the empty ring between r and R. Step 3 — Loop 2 lies entirely OUTSIDE the field region, so the flux through it is zero and stays zero: emf₂ = -d(0)/dt = 0. Answer: -(dB/dt)πr² in loop 1 and zero in loop 2 → option D.
NEET 2017
A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per metre. At the centre, a coil of 100 turns and radius 0.01 m is placed coaxially. The solenoid current reduces at a constant rate from 4 A to 0 A in 0.05 s. If the coil resistance is 10π² Ω, the total charge flowing through the coil is
A · 32π µC
B · 16 µC
C · 32 µC ✓
D · 16π µC
Solution: The changing solenoid current changes B, which induces an emf and current in the inner coil — a direct case of emf from a changing magnetic field. Step 1 — B inside solenoid: B = μ₀ n I = (4π×10⁻⁷)(2×10⁴)(4) = 32π×10⁻³ T at start, 0 at end, so ΔB = 32π×10⁻³ T. Step 2 — Charge q = ΔΦ_total / R = N × ΔB × A / R, where A = π(0.01)² = π×10⁻⁴ m² (coil area). Step 3 — q = [100 × 32π×10⁻³ × π×10⁻⁴] / (10π²) = (32π²×10⁻⁵)/(10π²) = 32×10⁻⁶ C = 32 µC → option C. Charge does not depend on the time 0.05 s — that is the neat trick.
Solved Electromagnetic Induction NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the electric field created in space by a magnetic field that is changing with time; it drives current in any loop placed there and exists even in empty space.
Is the induced electric field conservative or non-conservative?
Non-conservative. Its field lines are closed loops and ∮ E · dl = -dΦ/dt ≠ 0, so no single potential (voltage) can describe it, unlike the electrostatic field of charges.
What is the formula linking the induced E field and changing flux?
∮ E · dl = -dΦ/dt. For a circular region of radius r, inside (x < r): E = (x/2)(dB/dt); outside (x > r): E = (r²/2x)(dB/dt), which falls off as 1/x.
Can emf be induced without any battery or moving conductor?
Yes. A stationary loop in a changing magnetic field gets an emf purely from dB/dt. No motion (no motional emf) and no battery are needed — the changing field itself is the source.
Why does a loop outside the field region get zero emf?
emf depends on the flux enclosed by the loop. If the changing-B region is fully outside the loop, enclosed flux is zero, dΦ/dt = 0, so emf = 0, even though an induced E field is present along the loop.
