Final answer:
To find the radius of the solenoid, we used the formula for the self-inductance of a solenoid and rearranged it to solve for the radius. Using the given values, we determined that the radius of the solenoid is approximately 0.0222 meters, which corresponds to option (c).
Step-by-step explanation:
The student has asked to calculate the radius of an ideal solenoid that has 3000 turns, is 70.0 cm long, and has a self-inductance of 25.0 mH, given the permeability of free space μ0 = 4π ×10−7 T · m/A. To find the radius of the solenoid, we can use the formula for the self-inductance of a solenoid:
L = μ0 × (N2 × A) / l
Where L is the self-inductance, μ0 is the permeability of free space, N is the number of turns, A is the cross-sectional area of the solenoid (A=πr2 where r is the radius), and l is the length.
Assuming that end effects are negligible and rearranging the formula to solve for r, we get:
r = √((L × l) / (μ0 ×N2π))
By plugging in the given values:
r = √((25.0 × 10−3 H × 0.7 m) / (4π × 10−7 T · m/A × 30002π))
This yields:
r = √(17.5 × 10−6 / (4 × 9 × 106π))
r = √(17.5 × 10−6 / (36π × 106))
r = √(17.5 / (36π × 106))
r ≈ 0.0222 m
The radius of the solenoid is approximately 0.0222 m, which corresponds to option (c).