Question

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

• A. $10\times {10}^{-12}N$
• B. $8\times {10}^{-12}N$
• C. $2.5\times {10}^{-10}N$
• D. $8\times {10}^{-11}N$

Answer 1

The correct option is B $8\times {10}^{-12}N$

$\frac{1}{2}m{v}^2=2MeV\frac{1}{2}\times 1.6\times {10}^{-27}{v}^2=2\times {10}^6\times 1.6\times {10}^{-19}{v}^2=4\times {10}^{+14}v=2\times {10}^{+7}$
F=Bqv
$=2.5\times 1.6\times {10}^{-19}\times 2\times {10}^7=8\times {10}^{-12}N$