Question

A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many turns are required in a wire 2 long? The maximum permeability of the 4-79 permalloy is 200,000. The magnetic permeability of vacuum is = 410. (Enter your answer to three significant figures.)

Answer 1

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

`H = (nI)/(l)`

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

`n = (Hl)/(I)`

The magnetic field is again given by,

`H = (B)/(\mu_t)`

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

`n = ((B)/(\mu_t)l)/(I)`

Replacing with our values we have that,

`n = ((1.1Wb/m^2 )/(200000)(2m))/(4mA)`

`n = (((1.1Wb/m^2 )/(200000))((10^4 guass)/(1Wb/m^2))(2m))/(4mA((10^(-3)A)/(1mA)))`

`n = 27.5 \approx 28`

Therefore the number of turn required is 28Truns