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 follows a circular path. On a circular path, the velocity vector is al

Vedclass · Accepted Answer

$-1.5$

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(C) When a charged particle moves perpendicular to a uniform magnetic field, it follows a circular path. On a circular path, the velocity vector is always tangent to the path, and the acceleration vector (centripetal acceleration) is directed towards the center of the circle. Therefore, the velocity and acceleration vectors are always perpendicular to each other at any instant.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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