Velocity and acceleration vectors of a charge | vedclass

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(C) When a charged particle moves perpendicular to a uniform magnetic field,it undergoes uniform circular motion.In uniform circular motion,the accele

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$-1.5$

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(C) When a charged particle moves perpendicular to a uniform magnetic field,it undergoes uniform circular motion.In uniform circular motion,the acceleration (centripetal acceleration) is always directed towards the center of the circle,while the velocity vector is directed along the tangent to the path.Therefore,the velocity vector $\overrightarrow{v}$ and the acceleration vector $\overrightarrow{a}$ are always perpendicular to each other.Given,$\overrightarrow{v} = 2 \hat{i} + c \hat{j}$ and $\overrightarrow{a} = 3 \hat{i} + 4 \hat{j}$.Since $\overrightarrow{v} \perp \overrightarrow{a}$,their dot product must be zero:$\overrightarrow{v} \cdot \overrightarrow{a} = 0$$(2 \hat{i} + c \hat{j}) \cdot (3 \hat{i} + 4 \hat{j}) = 0$$(2)(3) + (c)(4) = 0$$6 + 4c = 0$$4c = -6$$c = -1.5$

Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is

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