Final answer:
The inductance (L) of a 0.60-m-long air-filled solenoid with a diameter of 2.9 cm containing 7500 loops can be calculated using the formula for a solenoid's inductance. After computing the cross-sectional area and substituting all values into the equation, the inductance is found to be approximately 0.087 Henry.
Step-by-step explanation:
To determine the inductance (L) of the 0.60-m-long air-filled solenoid with a diameter of 2.9 cm containing 7500 loops, we use the formula for the inductance of a solenoid:
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 (7500)
- A is the cross-sectional area (πr2, with r = 0.029/2 m)
- l is the length of the solenoid (0.60 m)
First, calculate the cross-sectional area:
A = π × (0.029/2)2 = 6.6079 x 10-4 m2
Now, substitute the values into the equation:
L = (4π x 10-7 H/m × 75002 × 6.6079 x 10-4 m2)/0.60 m
L = 0.087 H (rounded to three significant figures)
The inductance of the solenoid is approximately 0.087 Henry.