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9 April, 10:16

A wire with a linear mass density of 1.19 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic friction between the wire and the surface is 0.250. If the wire carries a current of 1.76 A westward and moves horizontally to the south, determine the magnitude and direction of the smallest magnetic field that can accomplish this.

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  1. 9 April, 11:04
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    The magnitude and direction of the smallest magnetic field that can accomplish this are 0.172 T and vertically to the east.

    Explanation:

    given information:

    density of wire, m/l = 1.19 g/cm = 0.119 kg/m

    coefficient of kinetic friction, μk = 0.250

    current, I = 1.7 A

    according to the Newton's first law

    ∑F = 0

    F (magnetic) - F (friction) = 0

    F (magnetic) = F (friction)

    F (magnetic) = B i l sin θ

    where

    F (magnetic) = magnetic force

    B = magnetic field (T)

    i = current (A)

    l = the length of wire (m)

    θ = angle

    and

    F (friction) = μk N

    where

    F (friction) = kinetic force (N)

    μk = coefficient of kinetic friction

    N = Normal force (N)

    so,

    F (magnetic) = F (friction)

    B i l sin θ = μk N

    B = μk N/i l sin θ

    now lets determine the normal force

    N - mg = 0

    N = mg

    therefore

    B = μk mg/i l sin θ

    θ = 90° because the magnetic field is perpendicular to the wire (smallest magnetic field)

    B = μk mg/i l sin 90

    = μk mg/i l

    = (μk g/i) (m/l)

    = ((0.250) (9.8) / (1.7)) ((0.119)

    = 0.172 T

    according to the right hand rule, the direction of the magnetic field is vertically to the east. (magnetic force to the south and current to the westward)
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