Final answer:
a. The energy stored in the given volume of the magnetic field is approximately 34.8 millijoules. b. The current in the wire windings of the solenoid is approximately 4.01 amperes.
Step-by-step explanation:
a. The energy stored in a magnetic field can be calculated using the formula E = (1/2)μ0B^2V, where E is the energy, μ0 is the permeability of free space (4π × 10^-7 T·m/A), B is the magnitude of the magnetic field, and V is the volume of the space. Plugging in the given values, we have E = (1/2)(4π × 10^-7 T·m/A)(49.9 × 10^-3 T)^2(24.6 × 10^-2 m)(π(1.82 × 10^-2 m)^2) ≈ 34.8 mJ. Therefore, the energy stored in the magnetic field is approximately 34.8 millijoules.
b. The inductance of a solenoid can be calculated using the formula L = (μ0N^2A)/l, where L is the inductance, μ0 is the permeability of free space, N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid. Rearranging the formula, we have N = sqrt((Ll)/(μ0A)). Plugging in the given values, we have N = sqrt((775 × 10^-6 H)(24.6 × 10^-2 m)/(4π × 10^-7 T·m/A)(π(1.82 × 10^-2 m)^2)) ≈ 98. Therefore, the number of turns in the solenoid is approximately 98. Given the relationship I = B·A·N, where I is the current, B is the magnetic field, A is the cross-sectional area, and N is the number of turns, we can rearrange the formula to solve for I. Plugging in the given values, we have I ≈ (49.9 × 10^-3 T)(π(1.82 × 10^-2 m)^2)(98) ≈ 4.01 A. Therefore, the current in the wire windings of the solenoid is approximately 4.01 amperes.