Question

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

• A. $8\times {10}^{-12}$N
• B. $4\times {10}^{-12}$N
• C. $16\times {10}^{-12}$N
• D. $2\times {10}^{-12}$N

Answer 1

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

$\frac{1}{2}{MV}^2=2MeV$

$V^2=\frac{2\times 2\times {10}^6\times 1.6\times {10}^{-19}}{1.65\times {10}^{-27}}$

$V^2=1.97\times {10}^{7}m/s$

$\frac{1.6\times 1.97\times 2.5\times {10}^{-19}\times {10}^7}{7.88\times {10}^{12}}$

$F=8\times {10}^{-12}$