Question

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the proton is

• A. $2.5\times {10}^{-10}N$
• B. $7.6\times {10}^{-11}N$
• C. $2.5\times {10}^{-11}N$
• D. $7.6\times {10}^{-12}N$

Answer 1

The correct option is D

$7.6\times {10}^{-12}N$

$F=qvB=1.6\times {10}^{-19}\times \left[\sqrt{\frac{2E}{m}}\right]2.5$
$=4\times {10}^{-19}\sqrt{\frac{2\times 2\times 1.6\times {10}^{-19}\times {10}^6}{1.66\times {10}^{-27}}}=7.6\times {10}^{-12}N$