Superconductors can carry very large currents with no resistance. If a superconducting wire is formed into a solenoid of length 20.0 cm with

Question

Superconductors can carry very large currents with no resistance. If a superconducting wire is formed into a solenoid of length 20.0 cm with 691 turns, what is the magnetic field inside the solenoid when the current is 5.38 kA? (µ0 = 4π × 10−7 T⋅m/A)

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Nguyệt Ánh 3 years 2021-08-21T19:21:56+00:00 1 Answers 19 views 0

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  1. thienhuong
    0
    2021-08-21T19:23:31+00:00
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    Answer:

    23.36 T

    Explanation:

    We are given that

    Length of solenoid=l=20 cm=\frac{20}{100}=0.2 m

    1 m=100 cm

    Number of turns=N=691

    Current=I=5.38kA=5.38\times 10^3 A

    1 kA=10^3 A

    \mu_0=4\pi\times 10^{-7}Tm/A

    We have to find the magnetic field inside the solenoid.

    We know that the magnetic field of solenoid

    B=\frac{\mu_0NI}{l}

    Substitute the values

    B=\frac{4\pi\times 10^{-7}\times 691\times 5.38\times 10^3}{0.2}

    B=23.36 T

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