Question

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

Answer 1

Answer:

M = 6.32 x 10⁻⁶ H

Explanation:

given,

Length of solenoid = 10 cm = 0.1 m

diameter = 0.40 cm

radius = 0.2 cm = 0.002

number of turns, N₁ = 800

                            N₂ = 50

mutual inductance will be equal to  

     $M = \dfrac{\mu_0N_1N_2A}{l}$

     $M = \dfrac{4\pi \times 10^{-7}\times 800\times 50 \times \pi \times (0.002)^2}{0.01}$

           M = 6.32 x 10⁻⁶ H

hence, mutual inductance of the combination of two coil is equal to  M = 6.32 x 10⁻⁶ H