“Some pacemakers employ magnetic reed switches to enable doctors to change their mode of operation without surgery. A typical reed switch ca

Question

“Some pacemakers employ magnetic reed switches to enable doctors to change their mode of operation without surgery. A typical reed switch can be switched from one position to another with a magnetic field of 5.0×10−4T. Part A What current must a wire carry if it is to produce a 5.0×10−4T field at a distance of 0.50 m?”

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Dâu 5 years 2021-08-03T05:45:59+00:00 1 Answers 48 views 0

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  1. Nho
    0
    2021-08-03T05:47:50+00:00
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    Answer:

    The current the wire must carry is 1250 A

    Explanation:

    Given;

    strength of magnetic field, B = 5.0 × 10⁻⁴ T

    distance from the wire, r = 0.50 m

    The strength of magnetic field on the current carrying wire is given as;

    B = \frac{\mu_o I}{2\pi r}

    where;

    B is the magnetic field strength

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

    I is the current on the wire

    r is the distance from the wire

    Make current “I” the subject of the formula;

    B = \frac{\mu_o I}{2\pi r} \\\\I = \frac{2\pi rB}{\mu_o} \\\\I = \frac{2\pi *0.5*5*10^{-4}}{4\pi *10^{-7} }\\\\I = 1.25 *10^3 \ A\\\\I = 1250 \ A

    Therefore, the current the wire must carry is 1250 A

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