A 7.00 x 10 -5 H solenoid is constructed by wrapping 66 turns of wire around a cylinder with a cross-sectional area of 6.2 x 10-4 m2. When the solenoid is shortened by squeezing the turns closer together, the inductance increases to 8.40 x 10-5 H. Determine the change in the length of the solenoid.
To determine the change in the length of the solenoid, we can use the formula for the inductance of a solenoid:
L = (μ₀ * N² * A) / l
where:
L is the inductance
μ₀ is the permeability of free space (constant)
N is the number of turns of wire
A is the cross-sectional area of the solenoid
l is the length of the solenoid
We can rearrange this formula to solve for l:
l = (μ₀ * N² * A) / L
Now that we have the formula, we can calculate the initial length of the solenoid using the given values:
l_initial = (μ₀ * N_initial² * A) / L_initial
where N_initial is the initial number of turns and L_initial is the initial inductance.
Then, we can calculate the final length of the solenoid using the given values:
l_final = (μ₀ * N_final² * A) / L_final
where N_final is the new number of turns and L_final is the new inductance.
Finally, we can find the change in length by subtracting the initial length from the final length:
change in length = l_final - l_initial
Now, let's plug in the given values:
μ₀ = 4π * 10^-7 T m/A
N_initial = 66 turns
A = 6.2 * 10^-4 m²
L_initial = 7.00 * 10^-5 H
N_final = 66 turns
L_final = 8.40 * 10^-5 H
Calculating these values, we get:
l_initial = (4π * 10^-7 T m/A) * (66 turns)² * (6.2 * 10^-4 m²) / (7.00 * 10^-5 H)
l_final = (4π * 10^-7 T m/A) * (66 turns)² * (6.2 * 10^-4 m²) / (8.40 * 10^-5 H)
change in length = l_final - l_initial
Calculating these equations will give us the change in length of the solenoid.