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An electron is travelling along the $x$ -direction. It encounters a magnetic field in the $y$ -direction. Its subsequent motion will be
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Verified Answer
The correct answer is:
a circle in the $x y$ -plane
If a particle carrying charge $q$ and moving with velocity $v$ through a point in magnetic field experiences a deflecting force $F$, then magnetic field $\overrightarrow{\mathbf{B}}$ is given by
Here,
$$
\begin{array}{l}
\overrightarrow{\mathbf{F}}=q \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}} \\
\overrightarrow{\mathbf{v}}=v_{x} \hat{\mathbf{i}} \text { and }
\end{array}
$$
$$
\overrightarrow{\mathbf{B}}=B_{y} \hat{\mathbf{j}}
$$
$\therefore$
$$
\overrightarrow{\mathbf{F}}=e v_{x} B_{y}(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=e v_{x} B_{y} \hat{\mathbf{k}}
$$
Hence, subsequent motion of the charged particle will be a circle in the $x y$ -plane.
Here,
$$
\begin{array}{l}
\overrightarrow{\mathbf{F}}=q \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}} \\
\overrightarrow{\mathbf{v}}=v_{x} \hat{\mathbf{i}} \text { and }
\end{array}
$$
$$
\overrightarrow{\mathbf{B}}=B_{y} \hat{\mathbf{j}}
$$
$\therefore$
$$
\overrightarrow{\mathbf{F}}=e v_{x} B_{y}(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=e v_{x} B_{y} \hat{\mathbf{k}}
$$
Hence, subsequent motion of the charged particle will be a circle in the $x y$ -plane.
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