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 $