Question

A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system

Answer 1

The mutual inductance is
`M = 0.000406 \ H`

Step-by-step explanation:

From the question we are told that

The number of turns per unit length is
`N = 1800`

The radius is
`r = 0.0165 \ m`

The number of turns of the solenoid is
`N_s = 210 \ turns`

Generally the mutual inductance of the system is mathematically represented as

`M = \mu_o * N * N_s * A`

Where A is the cross-sectional area of the system which is mathematically represented as

`A = \pi * r^2`

substituting values

`A = 3.142 * (0.0165)^2`

`A = 0.0008554 \ m^2`

also
`\mu_o` is the permeability of free space with the value
`\mu_o = 4\pi * 10^(-7) N/A^2`

So

`M = 4\pi * 10^(-7) *1800 * 210 * 0.0008554`

`M = 0.000406 \ H`