Demo Equipment

Notes: Faraday's law

Equation
  • ℰ = | ΔΦB | / Δt
  • Recall, T·m2 = V·s
  • → units of ℰ is Volts
  • Now, ℰ is not due to a battery.
  • Instead, it is due to the interaction between the charges in the conductor and the magnetic field.
Solenoids and coils
  • For a solenoid or coil, each loop has a flux.
  • Total flux through a solenoid = sum of each loop's flux.
  • For a uniform B field, ΦB, total = N ΦB, loop
  • ℰ = ΔΦB, total / Δt

Demo: Faraday's law

Observations
  • Voltmeter moves to positive as the coil covers the magnets, then to negative as the coil uncovers the magnets.
  • Magnitude of voltage? ~ 0.25 V
Repeat
  • Try 2 cm/s (should be ~ 1 V)
  • Try 8 cm/s (should be ~ 4 V)
Explanation
  • While covering...
  • ΦB, total = 400 (0.15 m × x) (0.10 T) = 6 T·m x
  • x = 4 cm/s × Δt
  • ΦB, total = 0.24 T·m2/s Δt
  • ℰ = 0.24 V
  • For 2 cm/s, ℰ will be cut in half (0.12 V).
  • For 8 cm/s, ℰ will double (0.48 V).

Long Problem: Faraday's law

Solution
  • 2: diagram
  • 3: L = 0.15 m, r = 0.010 m, Ninner = 250, Nouter = 1200, I = 1.2 A, Δt = 0.50 s
  • 4: principles - magnetic field of a solenoid, flux, Faraday's law
  • 5: B = μ0 Ninner I / L, ΦB, total = Nouter A B cos(θ), ℰ = ΔΦB, total / Δt
  • 6: Compute B due to inner, then change in flux in the outer, then ℰ
  • 7: B = 0.0025 T, ΔΦB, total = 0.00095 T·m2, &Esr; = 0.0019 V
  • 8: units (B → T, Φ → T·m2, ℰ → V). Solenoid field is < 1 T, which is reasonable.