Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductan

Question

Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is

8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

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Calantha 4 years 2021-07-28T01:38:15+00:00 1 Answers 10 views 0

Answers ( )

  1. Helga
    0
    2021-07-28T01:39:56+00:00
    Reply

    Answer:

    ∴Inductance of solenoid A is \frac18 of inductance of solenoid B.

    Explanation:

    Inductance of a solenoid is

    L=N\frac\phi I

     =N\frac{B.A}{I}

     =N\frac{\mu_0NI}{l.I}A

     =\frac{\mu_0N^2A}{l}

     =\frac{\mu_0N^2}{l}.\pi(\frac d2)^2

     =\mu_0\pi\frac{N^2d^2}{4l}

    N= number of turns

    l = length of the solenoid

    d= diameter of the solenoid

    A=cross section area

    B=magnetic induction

    \phi = magnetic flux

    I= Current

    Given that, Solenoid A has total number of turns N, length L and diameter D

    The inductance of solenoid A is

    =\mu_0\pi\frac{N^2D^2}{4L}

    Solenoid B has total number of turns 2N, length  2L and diameter 2D

    The inductance of solenoid B is

    =\mu_0\pi\frac{(2N)^2(2D)^2}{4.2L}

    =\mu_0\pi\frac{16 N^2D^2}{4.2L}

    Therefore,

    \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac{\mu_0\pi\frac{N^2D^2}{4L}}{\mu_0\pi\frac{16 N^2D^2}{4.2L}}

    \Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18

    \Rightarrow {\textrm{Inductance of A}}=\frac18\times {\textrm{Inductance of B}}

    ∴Inductance of solenoid A is \frac18 of inductance of solenoid B.

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