Question

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A>s. (a) What is the mutual inductance of these solenoids

Answer 1

Answer: The mutual inductance of these solenoids is
`2.88 * 10^(-7) H`.

Step-by-step explanation:

Given: Length = 50.0 cm (1 cm = 0.01 m) = 0.50 m

`N_(1)` = 6750

`N_(2)` = 15

Radius =
`(0.120 cm)/(2) = 0.6 cm = 6 * 10^(-4) m`

As inner of a solenoid resembles the shape of a circle. So, its area is calculated as follows.

`Area = \pi * r^(2) = \pi * (6 * 10^(-4))^(2)`

Formula used to calculate mutual conductance of two solenoids is as follows.

`M = (\mu_(o) * A * N_(1) * N_(2))/(l)`

where,

M = mutual conductance

A = area

`\mu_(o)` = relative permeability =
`4 \pi * 10^(-7) Tm/A`

`N_(1)` = no. of coils in outer solenoid

`N_(2)` = no. of coils in inner solenoid

l = length

Substitute the values into above formula as follows.

`M = (\mu_(o) * A * N_(1) * N_(2))/(l)\\= (4 \pi * 10^(-7) Tm/A * \pi (6 * 10^(-4))^(2) * 6750 * 15)/(0.5 m)\\= 2.88 * 10^(-7) H`

Thus, we can conclude that the mutual inductance of these solenoids is
`2.88 * 10^(-7) H`.