In a region of space where the magnetic field of the earth has a magnitude of 80 μT and is directed 30° below the horizontal, a 50-cm length

Question

In a region of space where the magnetic field of the earth has a magnitude of 80 μT and is directed 30° below the horizontal, a 50-cm length of wire oriented horizontally along an east-west direction is moved horizontally to the south with a speed of 20 m/s. What is the magnitude of the induced potential difference between the ends of this wire?

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Hồng Cúc 4 years 2021-08-07T19:23:45+00:00 1 Answers 54 views 0

Answers ( )

  1. thuthuy
    0
    2021-08-07T19:25:30+00:00
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    Answer:

    V_ind = 4 × 10^(-4) V

    Explanation:

    We are given;

    Magnetic field; B = 80 μT = 8 × 10^(-5) T

    Angle;θ = 30°

    Lenght;L = 50 cm = 0.5 m

    Speed; v = 20 m/s

    Now, formula for the induced potential difference is known as;

    V_ind = NBLVsin θ

    Where;

    V_Ind = Induced potential difference/voltage

    N = Number of turns

    B = Magnetic field

    V = velocity

    L = length

    Number of turns in this case is 1 since it’s just a wire between both ends.

    Thus, plugging in the relevant values, we have;

    V_ind = 1 × 8 × 10^(-5) × 20 × 0.5 × sin 30

    V_ind = 4 × 10^(-4) V

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