Q. A proton of velocity ($3\widehat{i} + 2\widehat{j}$)$\times 10^{5}ms^{- 1}$ enters a magnetic field $(2\widehat{i} + 3\widehat{k})$T. If the specific charge is $9.6 \times 10^{7}$C $kg^{- 1}$. The acceleration of the proton in $ms^{- 2}$ is
A) (6$\widehat{i} - 9\widehat{j} + 4\widehat{k}$)$\times 9.6 \times 10^{12}$
B) (6$\widehat{i} + 9\widehat{j} + 4\widehat{k}$)$\times 9.6 \times 10^{12}$
C) (6$\widehat{i} - 9\widehat{j} - 4\widehat{k}$)$\times 9.6 \times 10^{12}$
D) (6$\widehat{i} + 9\widehat{j} - 4\widehat{k}$)$\times 9.6 \times 10^{12}$
Explanation
(c) $q\left( \overrightarrow{v} \times \overrightarrow{B} \right) = \frac{mv^{2}}{r}$
or acceleration $= \frac{v^{2}}{r} = \frac{q}{m}\left( \overrightarrow{v} \times \overrightarrow{B} \right)$
$= 9.6 \times 10^{7} \times \lbrack\left( 3\widehat{i} + 2\widehat{j} \right)10^{5} \times (2\widehat{i} + 3\widehat{k})\rbrack$
$= 9.6 \times 10^{12} \times \left\lbrack 6\widehat{i} - 9\widehat{j} - 4\widehat{k} \right\rbrack ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.