A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.<

Question

A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?

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Ngọc Diệp 5 years 2021-08-11T01:37:05+00:00 1 Answers 198 views 1

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  1. thnahdat
    0
    2021-08-11T01:38:09+00:00
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    Answer:

    Explanation:

    Relation between flux and inductance is as follows

    φ = Li

    where φ is flux associated with induction of inductance L when a current i flows through it

    putting the values

    3.25 x 10⁻³ x 800 = L x 2.9

    L = .9 H

    for induced emf in an induction , the relation is

    emf induced = L di / dt

    Putting the values

    7.5 x 10⁻³ = .9 x di / dt

    di / dt = 8.33 x 10⁻³ A / s

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