Magnetic resonance imaging often need magnetic fields of a strength of around 1.50 T. The solenoid is 1.80 meters long and 75.0 cm in diamet

Question

Magnetic resonance imaging often need magnetic fields of a strength of around 1.50 T. The solenoid is 1.80 meters long and 75.0 cm in diameter. It is tightly wound with a single layer of 2.00 mm diameter superconducting wire.

Required:
What current is needed?

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Phúc Điền 4 years 2021-08-02T07:21:02+00:00 1 Answers 11 views 0

Answers ( )

  1. Giakhanh
    0
    2021-08-02T07:22:35+00:00
    Reply

    Answer:

    The current needed is 2387.32 A

    Explanation:

    Given;

    strength of the magnetic field, B = 1.5 T

    length of the solenoid, L = 18 m

    diameter of the solenoid, D = 75 cm = 0.75 m

    diameter of the superconducting wire, d = 2 mm = 0.002 m

    The number of turns of the solenoid is calculated as;

    N = \frac{length \ of \ solenoid}{diameter \ of \ wire } = \frac{1.8}{0.002} = 900 \ turns

    The magnetic field strength is given by;

    B = \frac{\mu_0 NI}{L} \\\\

    Where;

    I is the current needed

    μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

    I = \frac{BL}{\mu_0 N} =\frac{1.5 \times 1.8}{4\pi \times 10^{-7} \ \times 900} \\\\I = 2387.32 \ A

    Therefore, the current needed is 2387.32 A

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