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12 July, 17:18

A wire having a mass per unit length of 0.440 g/cm carries a 1.50-A current horizontally to the south. (a) What is the direction of the minimum magnetic field needed to lift this wire vertically upward?

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  1. 12 July, 17:33
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    B = 0.287 Tesla

    According to thumb rule the direction of magnet is to the east.

    Explanation:

    F = BiL sin∅

    note that F = mg

    B = magnetic field

    i = current

    therefore,

    mg = BiLsin∅

    Make B subject of the formula

    B = mg/iLsin∅

    m/L = mass per length of the wire and it is given in gram per centimeter. It should be converted to kg per meter.

    m/L = 0.440 g/cm

    convert to kg/m = 0.440 * 0.001 kg * 100 m = 0.044 kg/m

    B = 9.8 * 0.044 / sin 90 * 1.50 * 1

    B = 0.4312 / 1.50

    B = 0.28746666666

    B = 0.287 Tesla

    According to thumb rule the direction of magnet is to the east.
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