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Velocity and acceleration vectors of charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{\mathbf{v}}=2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}$ and $\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ respectively. Then the value of $c$ is
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$-1.5$
When a charged particle moves perpendicular to the magnetic field, it follows a circular path. On the circular path its velocity and acceleration will be at right angle to each other.
Given, $\overrightarrow{\mathbf{v}}=2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
As $\overrightarrow{\mathbf{v}}$ and $\overrightarrow{\mathbf{a}}$ are perpendicular, hence
$\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{a}}=0$
$(2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}) \cdot(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})=0 \Rightarrow 6+4 c=0 \Rightarrow c=-1.5$
Given, $\overrightarrow{\mathbf{v}}=2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
As $\overrightarrow{\mathbf{v}}$ and $\overrightarrow{\mathbf{a}}$ are perpendicular, hence
$\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{a}}=0$
$(2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}) \cdot(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})=0 \Rightarrow 6+4 c=0 \Rightarrow c=-1.5$
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