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A proton is projected with a velocity of $10^{7} m s^{- 1}$ at right angles to uniform magnetic field of induction 100 mT. The time taken by the proton to traverse $(6.25 + 10) = 16.25 %$ arc is (mass of proton = $1.6 \times 10^{- 27} k g$ , charge on proton = $1.6 \times 10^{- 19} C$ )

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$0.81 \times 10^{- 7} s$

$3.24 \times 10^{- 7} s$

$1.62 \times 10^{- 7} s$

$1 .0 5 \times 10^{- 7} s$

answer is C.

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Detailed Solution

$\begin{array}{l} t = \frac{T}{6} = \frac{2 π m}{6 \times B q} \\ = \frac{2 \times 3.14 \times 1.6 \times 10^{- 27}}{6 \times 10^{- 1} \times 1.6 \times 10^{- 19}} \end{array}$

$= 1.05 \times 10^{- 7} s$

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