A 0.26 mH inductor has a length that is four

times its diameter.
If it is wound with 34.4 cm−1
turns per
centimeter, what is its length?
Answer in units of m.

Assume that the length of the solenoid is much

larger than the solenoid’s radius and that the
core of the solenoid is air.
Calculate the inductance of a uniformly
wound solenoid having 260 turns if the length
of the solenoid is 25 cm and its cross-sectional
area is 5 cm2.
Answer in units of mH.

Calculate the self-induced emf in the solenoid
described in the first part if the current
through it is decreasing at the rate of 53 A/s.
Answer in units of mV

To find the length of the inductor, we can use the given information about its inductance, diameter, and winding density.

1. Start by converting the given inductance from millihenries to henries:
- 0.26 mH = 0.26 × 10^(-3) H

2. Now, let's find the number of turns in the inductor:
- The given winding density is 34.4 cm^(-1), which means that there are 34.4 turns for every centimeter of length.
- Since the inductor has a length that is four times its diameter, we can calculate the diameter:
- Let's assume the diameter is D, so the length is 4D.
- The circumference of a circle is given by C = πD. Since the inductor is wound around the diameter, the number of turns is equal to the circumference:
- Number of turns = 34.4 turns/cm × (πD)
- Substitute the value of D:
- Number of turns = 34.4 turns/cm × (π × D)
- And since the length is four times the diameter, we have:
- Number of turns = 34.4 turns/cm × (π × D) = 34.4 turns/cm × (π × (4D))
- Number of turns = 137.6 turns/cm × π × D

3. Now, we can use the formula for inductance to relate the number of turns, magnetic permeability, inductance, and the length of the inductor:
- Inductance (L) = (μ₀ × N² × A) / l
- L: inductance (given as 0.26 × 10^(-3) H)
- N: number of turns (calculated in the previous step as 137.6 turns/cm × π × D)
- A: cross-sectional area (assuming a circular cross-section, A = π(D/2)^2)
- l: length of the inductor (what we need to find)

4. Rearrange the formula to solve for the length (l):
- l = (μ₀ × N² × A) / L

By substituting the known values, we will be able to calculate the length of the inductor in meters.