Two solenoids have the same cross-sectional area and length, but the first one has twice as many turns er unit length as the second. <

Question

Two solenoids have the same cross-sectional area and length, but the first one has twice as many turns er unit length as the second.

The ratio of the self-inductance of the second solenoid to that of the first is :

a. 1:2

b. 1:4

c. 2:1

d. 4:1

e. 1:1

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Thiên Ân 3 years 2021-08-16T02:02:21+00:00 2 Answers 24 views 0

Answers ( )

  1. diemkieu
    0
    2021-08-16T02:03:59+00:00
    Reply

    Answer:

    option (b)

    Explanation:

    number of turns in the first solenoid, N1 = N

    number of turns in the second solenoid, N2 = 2N

    Let the current is i and the length is l and the area of crossection of the solenoid is A.

    The formula for the self inductance of the solenoid is

    L=\frac{\mu _{0}N^2\times A}{l}

    It means the self inductance of the solenoid is directly proportional to the square of number of turns.

    So, \frac{L_{1}}{L_{2}}=\frac{N^{2}}{4N^{2}}

    L1 : L2 = 1 : 4

  2. Jezebel
    0
    2021-08-16T02:04:02+00:00
    Reply

    Answer:

    b:1:4

    Explanation:

    We are given that two solenoid.

    Suppose ,the length of each solenoid=l

    Cross-sectional area of each solenoid=A

    Let , number of turns in in second  solenoid,N_2=N

    Number of turns in first solenoid,N_1=2 N

    We have to find the ratio of self-inductance of the second solenoid to that of the first.

    Self- inductance,L=\frac{\mu_0N^2A}{l}

    Using the formula

    Self- inductance of one solenoid,L_1=\frac{\mu_0(2N)^2A}{l}=\frac{\mu_04N^2A}{l}

    Self-inductance of second solenoid,L_2=\frac{\mu_0N^2A}{l}

    \frac{L_2}{L_1}=\frac{N^2}{4N^2}=\frac{1}{4}

    L_2:L_1=1:4

    Hence, option b is true.

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