Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of ${10}^{12} \mathrm{m}/{\mathrm{s}}^2$ by an applied magnetic field (West to east). The value of magnetic field: (Rest mass of proton is $1.6\times {10}^{-27}\mathrm{kg})$____mT. 

Given data:

Kinetic energy $1\mathrm{Mev}={10}^6\mathrm{ev}$

${10}^{6}ev=1.6\times {10}^{-13}$

Mass of proton $1.6\times {10}^{-27}$

Charge of proton$1.6\times {10}^{-19}$

Acceleration${10}^{12}\mathrm{m}/{\mathrm{s}}^2$

Explanation:

Kinetic energy=$½m{v}^2$

$1.6\times {10}^{-13}=½\times 1.6\times {10}^{-27}\times v^2$

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

Magnetic field

$Bqv=ma$

$B=\frac{ma}{qv}$

$B=\frac{1.6\times {10}^{-27}\times {10}^{12}}{1.6\times {10}^{-19}\times \sqrt{2}\times {10}^7}$

By solving we get

$B=0.71\times {10}^{-3}T$

$B=0.71 mT$