Question

A proton is projected with a velocity of $3\times {10}^{6}m{s}^{-1}$ perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.

Answer 1

$ma=qVB$

$a=\frac{qVB}{m}$

$=\frac{1.6\times {10}^{-19}\times 3\times {10}^6\times 0.6}{1.67\times {10}^{-27}}$

$=1.72\times {10}^{14}m/s^2$