Answer:
a) 1.26×10⁻⁴ H
b) 1.26×10⁻⁴ H
Step-by-step explanation:
N₁ = Number of turns in the solenoid = 600
N₂ = Number of turns in the coil = 30
A = Cross sectional area = 7×10⁻³ m²
l = Length of the solenoid = 0.4 m
μ₀ = Permeability of free space = 4π10⁻⁷ T m/A
a) Mutual inductance
[tex]M=\frac{\mu_{0} N_1N_2A}{l}\\\Rightarrow M=\frac{4\times 10^{-7}\times 600\times 30\times 7\times 10^{-3}}{0.4}\\\Rightarrow M=1.26\times 10^{-4}\ H[/tex]
∴ Mutual inductance of this system is 1.26×10⁻⁴ H
b) Mutual inductance does not depend on the cross sectional of the outer coil but depends on the area of the solenoid.
Hence, the mutual inductance remains the same.
