To find the cross-sectional area of the solenoid, we can use the formula for the energy stored in an inductor:
E = (1/2) * μ₀ * N² * A * I² / L
Where:
E is the energy stored in the solenoid,
μ₀ is the permeability of free space (approximately 4π × 10^(-7) T·m/A),
N is the number of turns per unit length,
A is the cross-sectional area of the solenoid,
I is the current flowing through the solenoid,
L is the length of the solenoid.
In this case, we are given:
E = 0.37 J,
N = 490 turns/m,
I = 11 A,
L = 1.1 m.
Rearranging the formula, we can solve for A:
A = (2 * E * L) / (μ₀ * N² * I²)
Substituting the given values:
A = (2 * 0.37 J * 1.1 m) / (4π × 10^(-7) T·m/A * (490 turns/m)² * (11 A)²)
Calculating this expression gives us:
A ≈ 0.00443 m²
Therefore, the cross-sectional area of the solenoid is approximately 0.00443 square meters.