A 3.6V light bulb is connected to a square loop of wire that measures 69cm per side. The loop is oriented perpendicular to a magnetic field with initial magnitude 2.2T. The magnetic field drops to zero over time
Δ
t. Find
Δ
t so that the bulb lights up with the specified potential difference.

Question

A 3.6V light bulb is connected to a square loop of wire that measures 69cm per side. The loop is oriented perpendicular to a magnetic field with initial magnitude 2.2T. The magnetic field drops to zero over time 
Δ
t. Find 
Δ
t so that the bulb lights up with the specified potential difference.

BlackTom AI · Accepted Answer

Faraday’s law of electromagnetic induction gives the magnitude of the induced emf: |ε| = |ΔΦ/Δt|, where magnetic flux Φ = B·A·cosθ. Because the square loop is perpendicular to the magnetic field, cosθ = 1 and Φ = B·A.

1) Relevant concepts:
- Magnetic flux Φ = B A (for perpendicular orientation).
- Faraday’s law: induced emf magnitude |ε| = |ΔΦ/Δt|.
- Use absolute change in flux when equating to the specified potential difference.

2) Step-by-step reasoning and calculations:
- Side length of the square loop: L = 69 cm = 0.69 m.
- Area A = L^2 = (0.69 m)^2 = 0.69 × 0.69 = 0.4761 m^2.
- Initial magnetic field B_i = 2.2 T, final B_f = 0, so ΔB = B_f − B_i = −2.2 T. The magnitude of the change is 2.2 T.
- Change in flux magnitude ΔΦ = A · |ΔB| = 0.4761 m^2 × 2.2 T.
  Intermediate multiplication: 0.4761 × 2 = 0.9522; 0.4761 × 0.2 = 0.09522; sum = 1.04742 Wb (webers).
- Required induced emf magnitude is ε = 3.6 V. From Faraday: ε = ΔΦ / Δt → Δt = ΔΦ / ε.
- Compute Δt = 1.04742 Wb / 3.6 V = 0.291505... s ≈ 0.29 s. Rounding to two significant figures or one decimal gives 0.3 s.

3) Why this answer is correct:
- The calculation directly applies Faraday’s law with correct area and field change; the numeric result ≈ 0.2915 s rounds to 0.3 s, matching the required potential difference.

4) Why the other choices are incorrect:
- 0.74 s or 0.4 s would give smaller induced voltages (ε = 1.04742/0.74 ≈ 1.42 V and ε = 1.04742/0.4 ≈ 2.62 V), both less than 3.6 V.
- 3.7 s yields ε ≈ 0.283 V, far too small.
- “None of these” is wrong because one option (0.3 s) matches the computed result after appropriate rounding.

Conclusion: The correct time interval is Δt ≈ 0.29 s, which rounds to 0.3 s, matching the provided answer 0.3s.

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