How to Use This Problem Set

This is your full workout for Electromagnetic Induction, grouped by theme: flux and Faraday's law, Lenz's-law direction calls, motional emf (sliding and rotating), mutual inductance, self-inductance and magnetic energy, and the AC generator.

Keep these handy:

  • ΦB=BAcos⁡θ\Phi_B = BA\cos\theta; ε=−NdΦBdt\varepsilon = -N\dfrac{d\Phi_B}{dt}; induced charge q=NΔΦRq = \dfrac{N\Delta\Phi}{R}
  • Motional: ε=Blv\varepsilon = Blv (sliding), ε=12BωR2\varepsilon = \frac{1}{2}B\omega R^2 (rotating)
  • Mutual: ε1=−M dI2dt\varepsilon_1 = -M\,\dfrac{dI_2}{dt}; coaxial solenoids M=μ0n1n2πr12lM = \mu_0 n_1 n_2 \pi r_1^2 l; concentric coils M=μ0πr122r2M = \dfrac{\mu_0\pi r_1^2}{2r_2}
  • Self: ε=−L dIdt\varepsilon = -L\,\dfrac{dI}{dt}; L=μ0n2AlL = \mu_0 n^2 A l; W=12LI2W = \frac{1}{2}LI^2; uB=B22μ0u_B = \dfrac{B^2}{2\mu_0}
  • Generator: ε=NBAωsin⁡ωt\varepsilon = NBA\omega\sin\omega t, ε0=NBAω\varepsilon_0 = NBA\omega
  • μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T m/A

Try each problem before reading its solution, and carry units through every step.

Solved Examples - Flux & Faraday's Law

Example 1. Flux through a loop of area 0.04 m2^2 in a 0.2 T field at 60 degrees to the normal:

Solution: Φ=BAcos⁡θ=0.2×0.04×0.5=4×10−3\Phi = BA\cos\theta = 0.2 \times 0.04 \times 0.5 = 4 \times 10^{-3} Wb.

Example 2. A 50-turn coil of area 2×10−32 \times 10^{-3} m2^2 sits normal to a field rising at 4 T/s. The emf?

Solution: ε=NAdBdt=50×2×10−3×4=0.4\varepsilon = NA\dfrac{dB}{dt} = 50 \times 2 \times 10^{-3} \times 4 = 0.4 V.

Example 3. (condensed) Square loop, side 10 cm, R=0.5 ΩR = 0.5\ \Omega, B=0.1B = 0.1 T at 45 degrees, killed in 0.7 s:

Solution: Φ0=10−32=7.07×10−4\Phi_0 = \dfrac{10^{-3}}{\sqrt 2} = 7.07 \times 10^{-4} Wb; ε=Φ0/0.7≈1.0\varepsilon = \Phi_0/0.7 \approx 1.0 mV; I=2I = 2 mA.

Example 4. (condensed) 500-turn coil, radius 10 cm, flipped 180 degrees in 0.25 s in BH=3×10−5B_H = 3\times10^{-5} T; R=2 ΩR = 2\ \Omega:

Solution: The change in flux per turn is ΔΦ=2BA=2×3×10−5×π×(0.1)2=1.885×10−6\Delta\Phi = 2BA = 2 \times 3\times10^{-5} \times \pi \times (0.1)^2 = 1.885\times10^{-6} Wb. Hence ε=NΔΦΔt=500×1.885×10−60.25≈3.8\varepsilon = \dfrac{N\Delta\Phi}{\Delta t} = \dfrac{500 \times 1.885\times10^{-6}}{0.25} \approx 3.8 mV; I=1.9I = 1.9 mA.

Example 5. ΦB=(2t2+5t+3)\Phi_B = (2t^2 + 5t + 3) mWb. EMF magnitude at t = 1 s?

Solution: ∣ε∣=∣dΦBdt∣=(4t+5)|\varepsilon| = \left|\dfrac{d\Phi_B}{dt}\right| = (4t+5) mV =9= 9 mV at t = 1 s.

Example 6. A 25-turn coil sees its per-turn flux change by 2×10−32 \times 10^{-3} Wb; circuit resistance 5 Ω\Omega. Charge through the circuit?

Solution: q=NΔΦR=25×2×10−35=10−2q = \dfrac{N\Delta\Phi}{R} = \dfrac{25 \times 2 \times 10^{-3}}{5} = 10^{-2} C — independent of how fast the change happened.

Example 7. Show the weber equals a volt-second.

Solution: From ε=dΦ/dt\varepsilon = d\Phi/dt, 1 Wb = 1 V s. (Also Wb = T m2^2 from Φ=BA\Phi = BA.)

Example 8. A 100-turn coil of area 10−210^{-2} m2^2 lies normal to a 0.4 T field and is pulled out of it in 0.2 s. Average emf?

Solution: Per-turn flux =0.4×10−2=4×10−3= 0.4 \times 10^{-2} = 4\times10^{-3} Wb \to 0. ε=100×4×10−30.2=2\varepsilon = \dfrac{100 \times 4\times10^{-3}}{0.2} = 2 V.

Solved Examples - Lenz's Law Directions

Example 9. A loop enters a field directed into the page. Induced current sense?

Solution: Into-page flux increasing → oppose with out-of-page flux → anticlockwise.

Example 10. The same loop later exits the region. Sense now?

Solution: Into-page flux decreasing → support it → clockwise.

Example 11. A magnet falls N-pole-down towards a horizontal ring. Current sense seen from above, and the force on the magnet?

Solution: Downward flux increasing → upper face becomes N → anticlockwise from above; the magnet is repelled (retarded).

Example 12. Why does a magnet dropped through a closed ring fall slower than g, but exactly at g through a cut ring?

Solution: Closed ring: induced current opposes the motion (approach and exit) → a < g. Cut ring: emf exists but no current, hence no opposing force → a = g.

Example 13. As the flux through a coil decreases, the induced current creates flux that:

Solution: Supports the dying flux (same direction) — opposition is always to the change, not to the flux.

Solved Examples - Motional EMF

Example 14. A 0.5 m rod moves at 4 m/s perpendicular to a 0.3 T field. EMF?

Solution: ε=Blv=0.3×0.5×4=0.6\varepsilon = Blv = 0.3 \times 0.5 \times 4 = 0.6 V.

Example 15. A jet with 25 m wingspan flies at 500 m/s where Bv=5×10−4B_v = 5\times10^{-4} T. Wing-tip emf?

Solution: ε=5×10−4×25×500=6.25\varepsilon = 5\times10^{-4} \times 25 \times 500 = 6.25 V.

Example 16. A 0.6 m rod rotates at 40 rad/s about one end, perpendicular to 0.5 T. EMF?

Solution: ε=12BωR2=12×0.5×40×0.36=3.6\varepsilon = \frac{1}{2}B\omega R^2 = \frac{1}{2} \times 0.5 \times 40 \times 0.36 = 3.6 V.

Example 17. Rails circuit: B = 0.4 T, l = 0.5 m, v = 3 m/s, R = 1.2 Ω\Omega. Find emf, current, applied force, and check the power balance.

Solution: ε=Blv=0.6\varepsilon = Blv = 0.6 V; I=0.5I = 0.5 A; F=BIl=0.4×0.5×0.5=0.1F = BIl = 0.4 \times 0.5 \times 0.5 = 0.1 N; Pmech=Fv=0.3P_{mech} = Fv = 0.3 W =I2R=0.25×1.2=0.3= I^2R = 0.25 \times 1.2 = 0.3 W. Balanced.

Example 18. In Example 17, which end of the rod is positive if v is rightward and B into the page (rod vertical)?

Solution: qv⃗×B⃗q\vec v\times\vec B points up the rod → the top end is positive — the rod is a battery of emf Blv.

Example 19. The rod now moves parallel to B. EMF?

Solution: Zero — no flux swept, v⃗×B⃗\vec v\times\vec B has no component along the rod.

Example 20. A wheel with 30 spokes of length R spins at ω\omega in field B. How does the axle-rim emf compare with a single-spoke rotor?

Solution: Identical: 12BωR2\frac{1}{2}B\omega R^2. The spokes are identical emf sources in parallel — the number is immaterial.

Solved Examples - Mutual Inductance

Example 21. Coaxial solenoids: n1=n2=500n_1 = n_2 = 500 turns/m, inner radius 5 cm, length 2 m. M?

Solution: M=μ0n1n2πr12l=4π×10−7×25×104×(π×2.5×10−3)×2≈4.9×10−3M = \mu_0 n_1 n_2 \pi r_1^2 l = 4\pi\times10^{-7} \times 25\times10^4 \times (\pi \times 2.5\times10^{-3}) \times 2 \approx 4.9 \times 10^{-3} H =4.9= 4.9 mH.

Example 22. M = 0.05 H; the neighbour's current changes at 20 A/s. Induced emf?

Solution: ∣ε∣=MdIdt=0.05×20=1.0|\varepsilon| = M\dfrac{dI}{dt} = 0.05 \times 20 = 1.0 V.

Example 23. Concentric coplanar coils: r1=2r_1 = 2 cm, r2=25r_2 = 25 cm. M?

Solution: M=μ0πr122r2=4π×10−7×π×4×10−40.5≈3.2×10−9M = \dfrac{\mu_0\pi r_1^2}{2r_2} = \dfrac{4\pi\times10^{-7} \times \pi \times 4\times10^{-4}}{0.5} \approx 3.2 \times 10^{-9} H.

Example 24. M = 1.5 H; current goes 0 to 20 A in 0.5 s. Flux-linkage change and average emf?

Solution: Δ(NΦ)=MΔI=30\Delta(N\Phi) = M\Delta I = 30 Wb; ε=1.5×40=60\varepsilon = 1.5 \times 40 = 60 V.

Example 25. Why is M12=M21M_{12} = M_{21} practically useful?

Solution: Compute whichever direction is easy (e.g. a long solenoid's uniform field through a small coil) and reciprocity hands you the hard direction free — a neat standard trick.

Solved Examples - Self-Inductance & Energy

Example 26. Solenoid: N = 500 turns, A = 10−310^{-3} m2^2, l = 0.25 m. L?

Solution: L=μ0N2Al=4π×10−7×2.5×105×10−30.25≈1.26×10−3L = \dfrac{\mu_0 N^2 A}{l} = \dfrac{4\pi\times10^{-7} \times 2.5\times10^5 \times 10^{-3}}{0.25} \approx 1.26 \times 10^{-3} H.

Example 27. L = 20 mH; current rises at 100 A/s. Back emf?

Solution: ∣ε∣=LdIdt=0.02×100=2|\varepsilon| = L\dfrac{dI}{dt} = 0.02 \times 100 = 2 V, opposing the rise.

Example 28. Energy in a 50 mH inductor at 4 A?

Solution: W=12LI2=12×0.05×16=0.4W = \frac{1}{2}LI^2 = \frac{1}{2} \times 0.05 \times 16 = 0.4 J.

Example 29. Magnetic energy density at B = 0.5 T?

Solution: uB=B22μ0=0.252×4π×10−7≈1.0×105u_B = \dfrac{B^2}{2\mu_0} = \dfrac{0.25}{2 \times 4\pi\times10^{-7}} \approx 1.0 \times 10^{5} J/m3^3.

Example 30. A solenoid's turn density n is tripled (same A, l). L changes by?

Solution: L∝n2L \propto n^2: factor 9.

Solved Examples - The AC Generator

Example 31. Kamla's coil: N = 100, A = 0.1 m2^2, B = 0.01 T, ν\nu = 0.5 rev/s. Peak voltage?

Solution: ε0=NBA(2πν)=100×0.01×0.1×π=0.314\varepsilon_0 = NBA(2\pi\nu) = 100 \times 0.01 \times 0.1 \times \pi = 0.314 V.

Example 32. N = 100, B = 0.2 T, A = 0.05 m2^2, ω\omega = 100 rad/s. Peak emf?

Solution: ε0=NBAω=100×0.2×0.05×100=100\varepsilon_0 = NBA\omega = 100 \times 0.2 \times 0.05 \times 100 = 100 V.

Example 33. For that generator, the emf when the coil plane is parallel to B?

Solution: Plane parallel to B means flux zero and changing fastest: ε=ε0=100\varepsilon = \varepsilon_0 = 100 V (maximum).

Example 34. B is doubled and omega halved. Peak emf and frequency?

Solution: ε0∝Bω\varepsilon_0 \propto B\omega: unchanged. Frequency ∝ω\propto \omega: halved. Peak same, slower alternation.