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zhuklara [117]
3 years ago
10

A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

ns 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.)
Physics
1 answer:
daser333 [38]3 years ago
4 0

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 = \frac{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 = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

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

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

Therefore the number of turn required is 28Truns

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