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Q.

A proton of kinetic energy 200 MeV moving towards east enters the magnetic field of $5 T .$ . If direction of field is from south to north and motion is upward, the force acting on it will be (mass of proton is $1.672 \times 10^{- 27} kg$ )

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a

zero

b

$1.6 \times 10^{- 10} N$

c

$3.2 \times 10^{- 8} N$

d

$1.6 \times 10^{- 6} N$

answer is B.

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Detailed Solution

$F = q v B$
But $\frac{1}{2} m v^{2} = K \Rightarrow v = \sqrt{\frac{2 K}{m}}$
$\begin{array}{l} \Rightarrow F = q \sqrt{\frac{2 K}{m}} B \\ = 1.6 \times 10^{- 19} \sqrt{\frac{2 \times 200 \times 1.6 \times 10^{- 13}}{1.67 \times 10^{- 27}}} \times 5 \\ ≅ 1.6 \times 10^{- 19} \times 2 \times 10^{8} \times 5 N = 1.6 \times 10^{- 10} N \end{array}$

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