Question

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. What is the magnetic force on the proton? $(M_p=1.6\times {10}^{-27}kg):$

• A. $8\times {10}^{-12}N$
• B. $4\times {10}^{-12}N$
• C. $6\times {10}^{-12}N$
• D. none of these

Answer 1

Answer: (A)

Magnetic force on proton is given by $F_m=q\upsilon B\sin \theta$ ...(i) (Here, $q=e=1.6\times {10}^{-19}C,B=2.5$ tesla) The kinetic energy of proton is $\frac{1}{2}{m}_p{\upsilon }^2=2MeV$ $=2\times {10}^6\times 1.6\times {10}^{-19}J$ So, $\upsilon =\frac{\sqrt{2\times 2\times {10}^6\times 1.6kt{10}^{-19}}}{1.6\times {10}^{-27}}$ $=2\times {10}^{7}m/s$ Now, from equation (i) we get $F_m=1.6\times {10}^{-19}\times 2\times {10}^7\times 2.5\sin {90}^o$ $=8\times {10}^{-12}N$