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Plz help me to finish this physics question?

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  • 天同
    Lv 7
    3 年前
    最愛解答

    (a) Given PQ = L, hence PR = L.cos(40), and RQ = L.sin(40)

    Area of loop = (1/2).(RQ).(PR) = (1/2).(L.sin(40)).(L.cos(40))

    = [(L^2)sin(40)cos(40)]/2

    Magnetic moment of current loop = current x area = [I.(L^2)sin(40)cos(40)]/2

    The direction of magnetic moment is given by the Right-Hand-Grip-Rule, and is pointing into the paper, i.e. in the -ve z-direction.

    (b) Torque = [B.I.(L^2)sin(40)cos(40)]/2

    The direction of torque is in the -ve y-direction.

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