One solenoid is centered inside another. The outer one has a length of   55.0 cmcm and contains 6750 coils, while the coaxial inner solenoid is 5.00 cmcm long and 0.190 cmcm in diameter and contains 17.0 coils. The current in the outer solenoid is changing at 35.5 A/sA/s .For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of The Tesla coil.Part AWhat is the mutual inductance of these solenoids?Express your answer in microhenrys.

To calculate the mutual inductance of the two solenoids, we need to use the formula for the mutual inductance of two solenoids:

M = (μ₀ * N₁ * N₂ * A₂) / l₁

where:

• μ₀ is the permeability of free space, which is 4π * 10^-7 T*m/A
• N₁ is the number of turns in the outer solenoid, which is 6750
• N₂ is the number of turns in the inner solenoid, which is 17
• A₂ is the cross-sectional area of the inner solenoid, which can be calculated using the formula for the area of a circle (π * r²), where r is the radius of the inner solenoid. The diameter of the inner solenoid is 0.190 cm, so the radius is 0.190/2 = 0.095 cm = 0.00095 m. Therefore, A₂ = π * (0.00095 m)²
• l₁ is the length of the outer solenoid, which is 55.0 cm = 0.55 m

Plugging these values into the formula, we get:

M = (4π * 10^-7 T*m/A * 6750 * 17 * π * (0.00095 m)²) / 0.55 m

After calculating the above expression, we get the mutual inductance in henrys. To convert this to microhenrys, we multiply by 10^6.

Please note that this is a simplified calculation. The actual mutual inductance may be slightly different due to factors such as the non-uniform magnetic field within the solenoids and the finite length of the inner solenoid.