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What current is required in the windings of a long solenoid that has 544 turns uniformly distributed over a length of 0.439 m in order to produce inside the solenoid a magnetic field of magnitude 0.0001132 T? The permeablity of free space is 1.25664 × 10−6 T m/A. Answer in units of mA.

Respuesta :

Answer:

I=72.7 mA

Explanation:

Given that

N= 544 turns

L= 0.439 m

Number of turn per unit length

n=N/L= 1238.17

B= 0.0001132 T

μo=1.25664 × 10⁻⁶

We know that magnetic filed given as'

B= μo n I

I =Current

[tex]I=\dfrac{B}{n\mu _o}[/tex]

By putting the values

[tex]I=\dfrac{0.0001132}{1238.17\times  1.25664\times 10^{-6}}[/tex]

I=0.0727

I=72.7 mA

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