A lightening bolt may carry a current of 11800 A for a short period of time. The permeability of free space is 1.25664 × 10−6 T · m/A. What

Question

A lightening bolt may carry a current of 11800 A for a short period of time. The permeability of free space is 1.25664 × 10−6 T · m/A. What is the resulting magnetic field 47 m from the bolt? Suppose that the bolt extends far above and below the point of observation.

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Thành Đạt 5 years 2021-07-18T09:13:30+00:00 1 Answers 31 views 0

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  1. Jezebel
    0
    2021-07-18T09:15:08+00:00
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    Answer:

    The resulting magnetic field is 5.021 x 10⁻⁵ T

    Explanation:

    Given;

    current in the lightening bolt, I = 11800 A

    distance from the bolt, r = 47 m

    permeability of free space, μ₀ = 1.25664 × 10⁻⁶ T· m/A

    Assume lightening bolt as long straight conductor, then the resulting magnetic field will be calculated as follows;

    B = \frac{\mu_oI}{2 \pi r}

    where;

    B is the resulting magnetic field

    I is the current in the bolt

    r is the distance from the bolt

    Substitute the given values and calculate B

    B = \frac{\mu_oI}{2 \pi r} \\\\B = \frac{1.25664 *10^{-6}*11800}{2 \pi (47)} \\\\B = 5.021 *10^{-5} \ T

    Thus, the resulting magnetic field is 5.021 x 10⁻⁵ T

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