Question

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

• A. 9.6× 10 ^15 N
• B. 7.6× 10 ^-12 N
• C. 9.6× 10 ^12 N
• D. 7.6× 10 ^12 N
  
  

Answer 1

The correct option is D 7.6× 10 ^12 N

We know that

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