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Question: A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the pr...

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the proton is

A

B

C

2.5×10−11N2.5 \times 10 ^ { - 11 } N

D

Answer

Explanation

Solution

F=qvB=1.6×10−19×[2Em]2.5F = q v B = 1.6 \times 10 ^ { - 19 } \times \left[ \sqrt { \frac { 2 E } { m } } \right] 2.5

=4×10−192×2×1.6×10−19×1061.66×10−27= 4 \times 10 ^ { - 19 } \sqrt { \frac { 2 \times 2 \times 1.6 \times 10 ^ { - 19 } \times 10 ^ { 6 } } { 1.66 \times 10 ^ { - 27 } } }