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A proton moving with a velocity , $2.5 \times 10^{7} m / s$ , enters a magnetic field of intensity 2.5 T making an angle 30° with the magnetic field. The force on the proton is

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$3 \times 10^{- 12} N$

$5 \times 10^{- 12} N$

$6 \times 10^{- 12} N$

$9 \times 10^{- 12} N$

answer is B.

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Detailed Solution

$F = qvBsinθ = 1.6 \times 10^{- 19} \times 2.5 \times 2.5 \times 10^{7} \sin 30^{o}$
$F = 1.6 \times 10^{- 19} \times 6.25 \times 10^{7} \times \frac{1}{2} = 5 \times 10^{- 12} N$

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