Final answer:
The total self-inductance of the heating coils in a hair dryer, assuming they act like a single solenoid, is approximately 6.28 x 10^-6 H.
Step-by-step explanation:
The student's question relates to the self-inductance of the heating coils in a hair dryer assumed to act like a single solenoid. Self-inductance is a property of a circuit (or a coil in this case) that opposes the change in current flowing through it. Given a solenoid with a certain number of turns, diameter, and length, we can calculate its self-inductance (L) using the formula:
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 per unit length of the solenoid, A is the cross-sectional area of the solenoid in square meters, and l is the length of the solenoid in meters.
In this example,
- Number of turns (N) = 400
- Diameter (d) = 0.800 cm (0.008 m radius)
- Length (l) = 1.00 m
- To find
- n
- , we divide the total number of turns by the length of the solenoid (n = N/l). The area
- A
- can be found using the formula for the area of a circle (A = π * r²). Substituting these values into the self-inductance formula, we get:
L = (4π x 10^-7 H/m) * (400/1.00 m)² * (π * (0.008 m)²) * 1.00 m
After calculating, we find that L ≈ 6.28 x 10^-6 H, which corresponds with option (b). This is the total self-inductance of the heating coils acting like a solenoid.