A long solenoid has a length of 0.54 m and contains 1900 turns of wire. There is a current of 2.5 A in the wire. What is the magnitude of th

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A long solenoid has a length of 0.54 m and contains 1900 turns of wire. There is a current of 2.5 A in the wire. What is the magnitude of the magnetic field within the solenoid?

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Diễm Thu 5 months 2021-08-28T11:29:38+00:00 1 Answers 2 views 0

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    2021-08-28T11:30:41+00:00

    Answer:

    Therefore,

    Strength magnetic field at within the Solenoid,

    B =1.05\times 10^{-2}\ T

    Explanation:

    Given:

    Turn = N = 1900

    length of solenoid = l = 0.54 m

    Current, I = 2.5 A

    To Find:

    Strength magnetic field within the Solenoid,

    B = ?

    Solution:

    If N is the number of turns in the length, the total current through the rectangle is NI. Therefore, Ampere’s law applied to this path gives

     \int {B} \, ds= Bl=\mu_{0}NI

    Where,

    B = Strength of magnetic field

    l = Length of solenoid

    N = Number of turns

    I = Current

    \mu_{0}=Permeability\ in\ free\ space=4\pi\times 10^{-7}\ Tm/A

    Therefore,

    B =\dfrac{\mu_{0}NI}{l}

    Substituting the values we get

    B =\dfrac{4\times 3.14\times 10^{-7}\times 1800\times 2.5}{0.54}= 1.05\times 10^{-2}\ T

    Therefore,

    Strength magnetic field at within the Solenoid,

    B = 1.05\times 10^{-2}\ T

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