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Many induction questions ask for the charge that flows, not the emf or the current. Students then try to find a time, get stuck because none is given, and lose the marks. The missing time is the clue: the charge does not depend on it. This post derives Q = ΔΦ/R, works three examples and shows where the shortcut stops working.
Where does Q = ΔΦ/R come from?
NCERT Class 12 Chapter 6, Electromagnetic Induction gives Faraday's law for a coil of N turns as ε = −N dΦ/dt. If the circuit's total resistance is R, the current is I = ε/R. Charge is current multiplied by time, added up over the whole change:
Q = ∫ I dt = −(N/R) ∫ (dΦ/dt) dt = −NΔΦ/R.
The minus sign only gives the direction (Lenz's law). The size is Q = NΔΦ/R. Time has cancelled. The CBSE 2026-27 curriculum and the JEE (Main) 2026 syllabus both list Faraday's law and induced emf and current; check your year's syllabus.
Worked example 1: flipping a coil
A coil of 100 turns, area 20 cm² = 2.0 × 10⁻³ m², sits with its plane perpendicular to a uniform field of 0.50 T. The circuit resistance is 10 Ω. The coil is flipped over (turned through 180°).
Flux per turn at the start: BA = 0.50 × 2.0 × 10⁻³ = 1.0 × 10⁻³ Wb.
After the flip the field passes through the other face, so the flux is −1.0 × 10⁻³ Wb. The change is 2BA = 2.0 × 10⁻³ Wb per turn.
Q = NΔΦ/R = 100 × 2.0 × 10⁻³/10 = 0.020 C.
If the coil is turned only through 90°, the flux goes from BA to zero and Q = 0.010 C. If it is pulled right out of the field, Q is also 0.010 C.
Worked example 2: does speed matter?
Take the coil from example 1 and pull it out of the field at a steady rate, first in 0.10 s, then in 0.010 s. ΔΦ total is NBA = 0.10 Wb.
Slow: emf = 0.10/0.10 = 1.0 V, current 0.10 A, charge 0.10 × 0.10 = 0.010 C.
Fast: emf = 0.10/0.010 = 10 V, current 1.0 A, charge 1.0 × 0.010 = 0.010 C.
Same charge. The heat produced is different, though: H = I²Rt gives 0.10² × 10 × 0.10 = 0.010 J for the slow pull and 1.0² × 10 × 0.010 = 0.10 J for the fast one. Pulling faster takes more work, which appears as more heat. Only the charge is independent of speed.
Worked example 3: measuring a field with a search coil
A search coil of 50 turns and area 10 cm² = 1.0 × 10⁻³ m² is connected to an instrument that records total charge; total resistance is 25 Ω. When the coil is flipped in an unknown field, the recorded charge is 4.0 × 10⁻⁴ C (sample numbers). Find B.
Q = 2NBA/R, so B = QR/(2NA) = (4.0 × 10⁻⁴ × 25)/(2 × 50 × 1.0 × 10⁻³) = 0.010/0.10 = 0.10 T.
This is why charge-measuring instruments can map a field without any stopwatch.
Does self-inductance change the answer?
If the circuit has self-inductance L, there is an extra term −L dI/dt. Integrated over the whole event it gives −LΔI. If the current is zero before and after, as it is once things settle, ΔI = 0 and the charge is still NΔΦ/R.
Why does this topic trip students up?
Zuza and colleagues (2014) note that electromagnetic induction is often taught quickly, with most time spent solving problems in a rote way, while research shows its basic concepts are barely understood by students. Q = ΔΦ/R is a good test of real understanding: you have to see that a formula with no time in it follows from one that has dΦ/dt.
Which mistakes cost marks?
Leaving out N. Flux linkage is NΦ.
Using area in cm². Convert to m² before multiplying.
Using only the coil's resistance. R is the whole circuit, including the meter.
Thinking heat, like charge, is independent of speed. Example 2 shows it is not.
A practice routine
Do one flip, one 90° turn and one removal problem with the same coil.
Repeat example 2 with your own times and confirm the charge stays fixed.
Do one search-coil problem, solving for B.
Then practise direction with our Faraday's and Lenz's law guide.
If you are unsure which flux change a problem means, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: find the charge when the coil in example 1 turns from 0° to 60°, where the flux falls from BA to BA cos 60°.
