Q. A proton (mass $= 1.67 \times 10^{- 27}kg$ and charge $= 1.6 \times 10^{- 19}C$) enters perpendicular to a magnetic field of intensity $2\ weber/m^{2}$ with a velocity $3.4 \times 10^{7}m/sec$. The acceleration of the proton should be
A) $6.5 \times 10^{15}\ m/\sec^{2}$
B) $6.5 \times 10^{13}\ m/\sec^{2}$
C) $6.5 \times 10^{11}\ m/\sec^{2}$
D) $6.5 \times 10^{9}\ m/\sec^{2}$
Explanation
(a) $F = ma = qvB \Rightarrow a = \frac{qvB}{m} = \frac{1.6 \times 10^{- 19} \times 2 \times 3.4 \times 10^{7}}{1.67 \times 10^{- 27}}$
$= 6.5 \times 10^{15}m/\sec^{2}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.