Final answer:
The induced current in the technician's wedding band is approximately 0.767 A.
Step-by-step explanation:
To calculate the induced current in the technician's wedding band, we can use Faraday's law of electromagnetic induction. The average emf induced in the ring can be found by multiplying the rate of change of magnetic flux with the number of turns in the ring.
The rate of change of magnetic flux can be calculated by multiplying the changing magnetic field strength with the area of the ring.
Given that the diameter of the ring is 2.2 cm, its radius is 1.1 cm and its cross-sectional area is
π * (1.1 cm)² = 3.801 cm² = 3.801 × 10⁻⁴ m².
Now, we can calculate the rate of change of magnetic flux:
ΔΦ = BΔA = (2.5 T - 0 T) × 3.801 × 10⁻⁴ m² = 9.5025 × 10⁻⁴ T.m²
Since the time taken for the field to change is given as 200 μs = 200 × 10⁻⁶ s and the ring is made of a gold alloy with resistivity 6.2 × 10⁻⁸ Ω·m, we can use Ohm's law V = IR to find the induced current I. Rearranging the formula, we have
I = V / R.
Substituting the values,
I = ΔΦ / (Δt * R) = 9.5025 × 10⁻⁴ T.m² / (200 × 10⁻⁶ s * 6.2 × 10⁻⁸ Ω·m).
Calculating this value, we find that the induced current in the technician's wedding band is approximately 0.767 A.