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Q.

A proton enters a magnetic field with a velocity of 2.5 x 107 m/s making angle 30° with the magnetic field. What is the force on the proton ? (Given B = 2.5T)

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a

1⋅25×10−12N

b

2⋅5×10−12N

c

5⋅0×10−12N

d

7⋅5×10−12N

answer is C.

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Detailed Solution

F=q∨Bsin⁡θ=1⋅6×10−192⋅5×107(2⋅5)sin⁡30∘
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