Question

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 $1.6 \times 10^{-27} kg )$

• A. $3 \times 10^{-12}N$
• B. $8 \times 10^{-10}N$
• C. $8 \times 10^{-12}N$
• D. $2 \times 10^{-10}N$
• E. $3 \times 10^{-10}N$

Answer 1

Answer: (C)

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