Final answer:
The self-inductance of the precision laboratory resistor coil is 176 millihenries.
Step-by-step explanation:
The self-inductance of a coil can be calculated using the formula:
L = (μ₀ * N² * A) / l
Where L is the self-inductance, μ₀ is the permeability of free space (4π x 10^-7 T·m/A), N is the number of turns, A is the cross-sectional area of the coil, and l is the length of the coil.
In this case, the diameter of the coil is given as 1.5 cm, so the radius is 0.75 cm. The cross-sectional area of the coil can be calculated as follows:
A = π * r²
Substituting the values into the formula:
A = π * (0.75 cm)²
= 1.77 cm²
= 1.77 x 10^-4 m²
Now, we can calculate the self-inductance:
L = (4π x 10^-7 T·m/A) * (500 turns)² * (1.77 x 10^-4 m²) / (2.75 cm)
L = 1.76 x 10^-4 H
= 176 mH