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18 June, 19:37

Suppose a straight 1.00-mm-diameter copper (density = 8.9 x103 kg/m3) wire could just "float" horizontally in air because of the force due to the earth's magnetic field, which is horizontal, perpendicular to the wire, and of magnitude 5.0 x 10-5 t. (a) what current would the wire carry?

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  1. 18 June, 20:06
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    We are given

    diameter, d = 1.00 mm

    density, ρ = 8.9 x10^3 kg/m3

    magnetic field, B = 5.0 x 10^-5 T

    We are asked to determine the current that would be carried by the wire

    We use the formula

    B = I R / A

    where I is the current

    R is the resistance and R = πρd / A

    So,

    B = I (πρd / A) / A

    B = I πρd / A²

    and

    A is the area, A = πd²/4

    So,

    B = I πρd / (πd²/4) ²

    B = I πρd / (π² d⁴ / 16)

    B = 16 I ρ / πd²

    Substituting the given and solving for I

    5.0 x 10^-5 = 16 I (8.9 x10^3 kg/m3) / π (1 x 10^-3) ²

    I = 1.1 x 10^-15 A
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