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27 March, 09:22

A 11.1 cm 11.1 cm long solenoid contains 703 turns 703 turns and carries a current of 4.31 A 4.31 A. What is the strength B B of the magnetic field at the center of this solenoid?

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Answers (2)
  1. 27 March, 12:15
    0
    Given Information:

    Current = I = 4.31 A

    Number of turns = N = 703

    Length of solenoid = L = 11.1 cm = 0.11 m

    Required Information:

    Magnetic field = B = ?

    Answer:

    Magnetic field = 0.346 T

    Step-by-step explanation:

    The magnetic field (B) produced in a current (I) carrying long solenoid of length (L) is given by

    B = μ₀NI/L

    Where μ₀ is the permeability of free space and its value is 4πx10⁻⁷ T. m/A

    B = (4πx10⁻⁷*703*4.31) / *0.11

    B = 0.0346 T

    Therefore, a magnetic field of strength 0.0346 T would be produced at the center of this solenoid.
  2. 27 March, 12:48
    0
    Given that

    Length of solenoid

    L=11.1cm = 0.111m

    Number of turns

    N=703turns

    Current

    I=4.31A

    B=?

    Thus, the magnetic field in the core of a solenoid is directly proportional to the product of the current flowing around the solenoid and the number of turns per unit length of the solenoid

    B=μo•n•I

    Where,

    μo is constant and has a value of

    μo=4π*10^-7

    n is number of turn per unit length

    n=N/L

    n=703/0.111 = 6333.333 turns/m

    I is the current, and it is 4.31A in this case

    Then,

    B=μo•n•i

    B=4π*10^-7*6333.333*4.31

    B=0.0343T

    B=34.3mT

    The magnetic field at the center of the coil is 34.3mT
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