A charged particle is moving in a gravity free space in a uniform magnetic field. If at an instant of time velocity vector is V→=4i^-3j^m/s and acceleration vector is a→=3 i^ -bj^m/s2 then magnitude of acceleration at that instant is

# A charged particle is moving in a gravity free space in a uniform magnetic field. If at an instant of time velocity vector is $\stackrel{\to }{\mathrm{V}}=\left(4\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$ and acceleration vector is  then magnitude of acceleration at that instant is

1. A

$2\sqrt{3}\mathrm{m}/{\mathrm{s}}^{2}$

2. B

3. C

4. D

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### Solution:

Negative force $\stackrel{\to }{\mathrm{F}}=\mathrm{Q}\stackrel{\to }{\mathrm{V}}×\stackrel{\to }{\mathrm{B}}$

$\therefore$ acceleration $\stackrel{\to }{\mathrm{a}}=\frac{\stackrel{\to }{\mathrm{F}}}{\mathrm{m}}=\left(\frac{\mathrm{Q}}{\mathrm{m}}\right)\stackrel{\to }{\mathrm{V}}×\stackrel{\to }{\mathrm{B}}$

Hence $\stackrel{\to }{a}$ is perpendicular to $\stackrel{\to }{V}$  Register to Get Free Mock Test and Study Material

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