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 $\vec{V} = (4 \hat{i} - 3 \hat{j}) m / s$ and acceleration vector is $\vec{a} = (\overset{}{3 \hat{i}} - b \hat{j}) m / s^{2}$ then magnitude of acceleration at that instant is

A
$2 \sqrt{3} m / s^{2}$
B
${4 m/s}^{2}$
C
$3 \sqrt{2} m / s^{2}$
D
${5 m/s}^{2}$

Solution:

Negative force $\vec{F} = Q \vec{V} \times \vec{B}$

$∴$ acceleration $\vec{a} = \frac{\vec{F}}{m} = (\frac{Q}{m}) \vec{V} \times \vec{B}$

Hence $\vec{a}$ is perpendicular to $\vec{V}$

$∴ \vec{a} . \vec{V} = 0 \Rightarrow 12 - 3 \times b = 0 \Rightarrow b = 4 ∴ |\vec{a}| = \sqrt{3^{2} + 4^{2}} m / s^{2} = 5 m / s^{2}$

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