Q. An $\alpha$-particle of mass $6.65 \times 10^{- 27}$ kg travels at right angles to a magnetic field of 0.2 T with a speed of $6 \times 10^{5}\ ms^{- 1}$. The acceleration of $\alpha$-particle will be
A) $5.77 \times 10^{11}\ ms^{- 2}$
B) $7.55 \times 10^{11}\ ms^{- 2}$
C) $5.77 \times 10^{12}\ ms^{- 2}$
D) $7.55 \times 10^{12}\ ms^{- 2}$
Explanation
(c) Acceleration, $a = \frac{F}{m} = \frac{qvB}{m}$
$= \frac{2 \times 1.6 \times 10^{- 19} \times 6 \times 10^{5} \times 0.2\ }{6.65 \times 10^{- 27}}$
$= 5.77 \times 10^{12}ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.