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Physics Question on Magnetic Field

A proton with energy of 2 MeV enters a uniform magnetic field of2.5 T normally. The magnetic force on the proton is (Take mass of proton to be 1.6×10−27kg)1.6 \times 10^{-27} kg )

A

3×10−12N3 \times 10^{-12}N

B

8×10−10N8 \times 10^{-10}N

C

8×10−12N8 \times 10^{-12}N

D

2×10−10N2 \times 10^{-10}N

Answer

8×10−12N8 \times 10^{-12}N

Explanation

Solution

Energy of proton = 2 MeV
\hspace15mm =2 \times 1.6\times10^{-19}\times10^6 J
\hspace15mm =3.2 \times 10^{-13}J
Magnetic field (B)=2.5 T
Mass of proton (m) =1.6×10−27kg=1.6 \times 10^{-27}kg
Energy of proton E=12mv2E= \frac {1}{2}mv^2
\therefore \hspace5mm v= \sqrt {\frac {2E}{m}} \hspace10mm ...(i)
Magnetic force on proton
\hspace15mm F=Bqv \, sin \, 90^\circ
\hspace15mm =Bqv
Substituting the value of v from E (i)
\hspace15mm F=Bq \sqrt {\frac {2E}{m}}
\hspace15mm =2.5 \times 1.6\times10^{-19} \sqrt {\frac {2\times3.2\times10^{-13}}{1.6 \times 10^{-27}}}
\hspace15mm =8 \times 10^{-12}N