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EAPCET Physics · Intermediate second year

Electromagnetic Induction

Electromagnetic Induction covers Faraday’s and Lenz’s laws, motional emf, self and mutual inductance, and the AC generator.

Subject
Physics
Taught in
Intermediate second year
Physics in the paper
40 of 160 questions

Key formulas and results

  • ε=−dΦdt\varepsilon = -\frac{d\Phi}{dt}
  • Motional emf: ε=Blv\varepsilon = Blv
  • Self-induced emf: ε=−LdIdt\varepsilon = -L\frac{dI}{dt}; energy stored 12LI2\tfrac{1}{2}LI^{2}
  • Solenoid inductance: L=μ0n2AlL = \mu_{0}n^{2}Al
  • Rotating coil: ε=NBAωsin⁡ωt\varepsilon = NBA\omega\sin\omega t
  • Flux through a coil of NN turns: Φ=NBAcos⁡θ\Phi = NBA\cos\theta
  • Mutual inductance: M=kL1L2M = k\sqrt{L_{1}L_{2}}, at most L1L2\sqrt{L_{1}L_{2}} when k=1k = 1
  • Growth of current in an LR circuit: i=i0(1−e−t/τ)i = i_{0}\left(1 - e^{-t/\tau}\right), with τ=L/R\tau = L/R
  • Charge that flows through a circuit when flux changes: Δq=ΔΦ/R\Delta q = \Delta\Phi/R (independent of how fast it changes)
  • Rod of length ll rotating about one end in a perpendicular field: ε=12Bωl2\varepsilon = \tfrac{1}{2}B\omega l^{2}
  • Flux through a rectangular loop (side LL parallel to a long wire, width bb, near side at distance aa): Φ=μ0IL2πln⁡a+ba\Phi = \frac{\mu_{0}IL}{2\pi}\ln\frac{a + b}{a}
  • Induced electric field inside a region of changing uniform BB (radius RR): E=r2⋅dBdtE = \frac{r}{2}\cdot\frac{dB}{dt} for r<Rr < R
  • Rod on rails with resistance RR: I=BlvRI = \frac{Blv}{R}, retarding force F=B2l2vRF = \frac{B^{2}l^{2}v}{R}, power needed P=B2l2v2RP = \frac{B^{2}l^{2}v^{2}}{R}
  • LR circuit: growth i=VR(1−e−Rt/L)i = \frac{V}{R}\left(1 - e^{-Rt/L}\right); decay i=i0e−Rt/Li = i_{0}e^{-Rt/L}; time constant τ=L/R\tau = L/R
  • Mutual inductance M=kL1L2M = k\sqrt{L_{1}L_{2}} with 0≤k≤10 \le k \le 1; small coil (N2N_{2} turns, radius bb) at the centre of a large coil (N1N_{1} turns, radius aa): M=μ0N1N2πb22aM = \frac{\mu_{0}N_{1}N_{2}\pi b^{2}}{2a}

What this chapter covers

  • Magnetic flux
  • Faraday’s law
  • Lenz’s law
  • Motional emf
  • Self and mutual inductance
  • AC generator

How it is usually asked

  • Induced emf from changing flux
  • Direction of induced current
  • Inductance and stored energy

Understanding Electromagnetic Induction

Magnetic flux through a coil is Φ = BA cos θ, where θ is the angle between the field and the normal to the coil. Faraday’s law says an emf is induced whenever this flux changes, whether because B changes, the area changes, or the coil rotates; for N turns the emf is −N dΦ/dt. Lenz’s law, the minus sign, says the induced current always opposes the change that causes it, which is energy conservation in action.

A conductor of length l moving with speed v perpendicular to a field B has a motional emf Blv. A changing current in a coil induces an emf in the same coil (self-inductance L) or in a nearby coil (mutual inductance M), and an inductor stores energy ½LI². A coil rotating in a uniform field produces a sinusoidal emf NBAω sin ωt, the basis of the AC generator.

EAPCET questions typically ask for induced emf, the direction of the induced current, inductance or stored energy, and the peak emf of a generator. Remember that an emf appears only while the flux is changing: a closed loop moving through a uniform field has no induced current because its flux stays constant, and a steady current in a primary coil induces nothing in a nearby secondary coil.

Worked example

EAPCET-style question

A coil of 100 turns and area 0.02 m20.02\ \text{m}^{2} is held perpendicular to a uniform magnetic field. The field is reduced from 0.5 T to 0.1 T in 0.2 s. The magnitude of the average induced emf is

  1. (1)4 V
  2. (2)0.04 V
  3. (3)6 V
  4. (4)0.4 V

Solution

  1. Change in flux per turn: ΔΦ=A ΔB=0.02×(0.5−0.1)=0.008 Wb\Delta\Phi = A\,\Delta B = 0.02 \times (0.5 - 0.1) = 0.008\ \text{Wb}.
  2. Average emf =NΔΦΔt=100×0.0080.2= N\frac{\Delta\Phi}{\Delta t} = \frac{100 \times 0.008}{0.2}.
  3. Average emf = 4 V.

Answer: 4 V

Shortcuts and results to remember

  • A rod sliding on rails reaches terminal velocity when the magnetic force BIlBIl balances the driving force; for a rod falling on smooth vertical rails closed by RR, v=mgRB2l2v = \frac{mgR}{B^{2}l^{2}}.
  • Motional emf is zero when the velocity is parallel to the rod or to B; for a bent rod, use the straight line joining its ends as the effective length.
  • Turning a loop through 180° in a uniform field changes its flux by 2BA2BA, not zero.
  • A loop moving entirely inside a uniform field has zero net emf and no current, although each side perpendicular to vv still has an emf vBlvBl; current flows only while it enters or leaves the field.
  • An inductor acts like an open switch at t=0t = 0 and like a plain wire at t=∞t = \infty in a DC circuit.
  • By Lenz’s law a magnet dropped through a closed ring is repelled, so it falls with acceleration less than gg and reaches the ground with 12mv2<mgh\tfrac{1}{2}mv^{2} < mgh.

More practice questions

Practice question 1

A square coil of 200 turns and side 18 cm has resistance 2 Ω. A magnetic field perpendicular to its plane rises uniformly from 0 to 0.5 T in 0.8 s. The induced current in the coil is

  1. (1)1.01 A
  2. (2)2.025 A
  3. (3)4.05 A
  4. (4)0.81 A

Solution

  1. Area A=0.18×0.18=0.0324 m2A = 0.18 \times 0.18 = 0.0324\ \text{m}^{2}
  2. ε=NAΔBΔt=200×0.0324×0.50.8\varepsilon = NA\frac{\Delta B}{\Delta t} = \frac{200 \times 0.0324 \times 0.5}{0.8}
  3. ε=4.05 V\varepsilon = 4.05\ \text{V}
  4. I=εR=4.052=2.025 AI = \frac{\varepsilon}{R} = \frac{4.05}{2} = 2.025\ \text{A}

Answer: 2.025 A

Practice question 2

A square loop of area 20 cm220\ \text{cm}^{2} and resistance 5 Ω lies perpendicular to a uniform field of 2 T. It is rotated through 180° about one of its sides in 0.01 s. The charge that flows through the loop is

  1. (1)8×10−4 C8 \times 10^{-4}\ \text{C}
  2. (2)1.6×10−3 C1.6 \times 10^{-3}\ \text{C}
  3. (3)3.2×10−3 C3.2 \times 10^{-3}\ \text{C}
  4. (4)4×10−4 C4 \times 10^{-4}\ \text{C}

Solution

  1. Initial flux Φi=BA=2×20×10−4=4×10−3 Wb\Phi_{i} = BA = 2 \times 20 \times 10^{-4} = 4 \times 10^{-3}\ \text{Wb}; after 180° it is −4×10−3 Wb-4 \times 10^{-3}\ \text{Wb}
  2. ΔΦ=8×10−3 Wb\Delta\Phi = 8 \times 10^{-3}\ \text{Wb}
  3. Δq=ΔΦR=8×10−35=1.6×10−3 C\Delta q = \frac{\Delta\Phi}{R} = \frac{8 \times 10^{-3}}{5} = 1.6 \times 10^{-3}\ \text{C}
  4. The charge does not depend on the time taken; the 0.01 s only fixes the average emf (0.8 V) and current (0.16 A)

Answer: 1.6×10−3 C1.6 \times 10^{-3}\ \text{C}

Common mistakes to avoid

  • Forgetting to multiply by the number of turns N in Faraday’s law.
  • Using sin θ in Φ = BA cos θ when the angle is given between B and the plane of the coil — θ must be measured from the normal to the plane.
  • Adding the initial and final fields instead of taking their difference — only the change in flux induces an emf (unless the field reverses direction).
  • Getting the direction of the induced current wrong — it must oppose the change in flux, not the flux itself.

How to prepare it

For direction questions, ask what the induced current must do to oppose the change in flux.

Once the formulas feel familiar, test them the way the exam does: practise Electromagnetic Induction questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Electromagnetic Induction chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Electromagnetic Induction part of the EAPCET Physics syllabus?
Yes. Electromagnetic Induction is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Electromagnetic Induction for EAPCET?
For direction questions, ask what the induced current must do to oppose the change in flux.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Electromagnetic Induction?
Forgetting to multiply by the number of turns N in Faraday’s law. Using sin θ in Φ = BA cos θ when the angle is given between B and the plane of the coil — θ must be measured from the normal to the plane. Adding the initial and final fields instead of taking their difference — only the change in flux induces an emf (unless the field reverses direction). Getting the direction of the induced current wrong — it must oppose the change in flux, not the flux itself.