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Most induction problems ask for an emf in a loop of wire. A harder question asks what happens where there is no wire at all: if the current in a solenoid is increasing, is there an electric field in the empty space inside? There is, and it goes round in circles. This post explains where that field comes from, shows how to calculate it inside and outside a solenoid, works an example with real numbers, and lists the mistakes students make when they treat it like the field of a charge.
Where does this come from in NCERT?
The current NCERT Class 12 Physics Part I chapter Electromagnetic Induction (2026-27 reprint) raises the question directly in Section 6.6. For a conductor at rest in a changing magnetic field, the force on its charges is F = q(E + v × B) = qE, because v = 0. So the force, and the induced emf, must come from an electric field, and NCERT concludes that "a time-varying magnetic field generates an electric field". It adds that such fields have properties different from those produced by static charges.
NCERT stops there and does not calculate the field. The JEE (Main) 2026 syllabus lists "Faraday's law, induced emf and current, Lenz's law" and does not name the induced electric field separately, so treat the calculation below as Faraday's law applied with symmetry, not as a new formula. The NEET (UG) 2026 syllabus lists Faraday's law too. Check your own exam year's syllabus.
How is this electric field different?
The field of a charge starts on positive charges and ends on negative ones. If you carry a test charge round any closed path in it, the total work is zero, which is why it has a potential.
The field made by a changing B has no charges to start or end on. Its lines close on themselves, like the field lines of B. Carry a charge once round a loop and the work done per unit charge is the emf:
∮E·dl = −dΦ/dt
This is Faraday's law with the wire removed. A wire only gives the charges a path to move along; the field is there either way.
How do you find E inside a solenoid?
Take a long solenoid of radius R with n turns per metre, carrying current I. Inside, B = μ0nI along the axis, and outside it is close to zero. If I changes, dB/dt = μ0n(dI/dt).
By symmetry, E must have the same size at every point on a circle of radius r centred on the axis, and it must point along that circle. Choose that circle as the loop.
Left side: ∮E·dl = E × 2πr.
Right side, inside (r < R): the loop encloses flux πr²B, so the emf is πr²(dB/dt).
So E × 2πr = πr²(dB/dt), giving E = (r/2)(dB/dt).
Outside (r > R), the loop still encloses only the flux in the solenoid, πR²B. So E × 2πr = πR²(dB/dt), giving E = (R²/2r)(dB/dt).
The field is zero on the axis, rises in proportion to r up to the windings, and then falls as 1/r. Lenz's law sets its direction: the field drives current that would oppose the change in flux. For the direction rules, see our guide to Lenz's law and induction problems.
Worked example: how large is the field?
A long solenoid has 2000 turns per metre and radius 2.0 cm. Its current rises steadily at 50 A/s. Find E at r = 1.0 cm, at the windings, and at r = 5.0 cm.
dB/dt = μ0n(dI/dt) = (4π × 10⁻⁷)(2000)(50) = 0.126 T/s.
Inside, r = 0.010 m: E = (0.010/2)(0.126) = 6.3 × 10⁻⁴ V/m.
At the windings, r = 0.020 m: E = (0.020/2)(0.126) = 1.26 × 10⁻³ V/m, the largest value.
Outside, r = 0.050 m: E = (0.020²)(0.126)/(2 × 0.050) = 5.0 × 10⁻⁴ V/m.
Check step 2 another way: a ring of radius 1.0 cm would have emf π(0.010)²(0.126) = 3.95 × 10⁻⁵ V. Dividing by the ring's length, 2π(0.010) = 0.0628 m, gives 6.3 × 10⁻⁴ V/m again.
Note one more result: every loop outside the solenoid that goes round it once has the same emf, π(0.020)²(0.126) = 1.58 × 10⁻⁴ V, however large the loop.
Which mistakes cost marks?
Looking for charges. There are none. Do not try Coulomb's law or Gauss's law for this E.
Using the area of your loop outside. Outside, only the solenoid's cross-section carries flux, so use πR², not πr².
Saying E = 0 where B = 0. E depends on enclosed changing flux, as the callout says.
Using a potential difference. Because ∮E·dl is not zero, this field has no single-valued potential. Speak of emf round a loop, not voltage between two points.
Forgetting the sign. The minus sign in Faraday's law fixes the direction; find it with Lenz's law, then use magnitudes.
For a similar emf with motion instead of a changing field, compare our post on motional emf in a rod on rails. If you find a field you cannot see hard to picture, Fermi's Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Sketch E against r for the worked example, marking the peak at r = R.
Repeat the example with the current falling at 50 A/s and state the new direction using Lenz's law.
Find the emf in a square loop of side 6 cm centred on the solenoid's axis. (Hint: the flux is still that of the solenoid alone.)
Write, in one sentence, why this field has no potential. If you can explain it to a friend, you understand it.
Today, derive E = (r/2)(dB/dt) from a blank page and check it against the worked example.
