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A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T · m2 enclosed by one turn of the coil (note that

Φ = kI,

and you can calculate the proportionality constant k). Determine the inductance in henries.

Respuesta :

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, [tex]\phi=5\times 10^{-3}\ Tm^2[/tex]

The relation between the self inductance and the magnetic flux is given by :

[tex]L=\dfrac{N\phi}{I}[/tex]

[tex]L=\dfrac{17\times 5\times 10^{-3}}{4}[/tex]

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

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