A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is  $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$
  • A$0.81\times10^{-7}$
  • B$1.62\times10^{-7}$
  • C$2.43\times10^{-7}$
  • D$3.24\times10^{-7}$
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