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A proton moving with a velocity, $2.5 \times 10^{5} m / \sec$ , enters a magnetic field of intensity 2.5T making an angle $30^{0}$ with the magnetic field. The force on the proton is

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$3 \times 10^{- 14} N$

$5 \times 10^{- 14} N$

$6 \times 10^{- 14} N$

$9 \times 10^{- 14} N$

answer is B.

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Detailed Solution

$v = 2.5 \times 10^{7} m / s$

$B = 2.5 T,$

$θ = 30^{0},$

$q = 1.6 \times 10^{- 19} C$

$F = ?$

$F = B q v \sin θ$

$F = (2.5) (1.6 \times 10^{- 19}) (2.5 \times 10^{5}) (\sin 30)$

$F = 6.25 \times 1.6 \times (1 / 2) \times 10^{- 14}$

$= 6.25 \times 0.8 \times 10^{- 14}$

$F = 5 \times 10^{- 14} N$

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