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Physics Question on Moving charges and magnetism

A proton of energy 2 MeV is moving perpendicular to uniform magnetic field of 2.5 T. The force on the proton is (Mp=1.6×1027 Kg(Mp = 1.6 \times 10^{27}\, Kg and qp=e=1.6×10−19C)q_p = e = 1.6 \times 10^{-19}C)

A

2.5×10−16 N2.5 \times 10^{-16}\,N

B

8×10−11 N8 \times 10^{-11}\,N

C

2.5×10−11 N2.5 \times 10^{-11}\,N

D

8×10−12 N8 \times 10^{-12}\,N

Answer

8×10−12 N8 \times 10^{-12}\,N

Explanation

Solution

F=Bqv=Bq2KEm(∵12mv2=KE)F=Bqv =Bq \sqrt{\frac{2KE}{m}}(\because \frac{1}{2}mv^2 =KE)