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A ^9 {\rm {\rm Be}} nucleus containing four protons and five neutrons has a mass of 1.50×10−26 {\rm kg} and is traveling vertically upward at 1.45 {\rm km}/{\rm s}.
If this particle suddenly enters a horizontal magnetic field of 1.15 {\rm T} pointing from west to east, find the magnitude of its acceleration vector the instant after it enters the field.

  • Physics -

    m = 9•1.66•10^-27 kg
    q = 4•1.6•10^-19 C
    ma = q• v• B•sinα
    sinα =1, v =1.45 km/s = 1450 m/s,
    B = 1.15 mT =1.15•10^-3 T,
    a = q• v• B/m =
    4•1.6•10^-19•1450•1.15•10^-3/9•1.66•10^-27 = 7.15•10^7 m/s^2

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