Final answer:
The self-inductance of a precision laboratory resistor made of a coil of wire with a 1.50 cm diameter, 4.00 cm length, and 500 turns is approximately 2.25 x 10^-6 H, which matches option (a). This calculation uses the formula for the self-inductance of a long solenoid and includes the permeability of free space, number of turns, cross-sectional area, and length of the solenoid.
Step-by-step explanation:
Calculating the Self-Inductance
The formula for the self-inductance (L) of a long solenoid is given by:
L = (μ0 N2 A) / l
Where:
- μ0 is the permeability of free space (4π x 10-7 H/m)
- N is the number of turns (500 turns)
- A is the cross-sectional area of the solenoid (πr2)
- l is the length of the solenoid (4.00 cm = 0.04 m)
Given that the diameter (d) is 1.50 cm, the radius (r) is 0.75 cm = 0.0075 m:
A = πr2 = 3.14159 * (0.0075 m)2
L = (4π x 10-7 H/m * 5002 * π * (0.0075 m)2) / 0.04 m
Upon calculating, we find that the self-inductance of the coil is approximately 2.25 x 10-6 H, which corresponds to option (a).