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An electron is moving in a circle of radius \( r \) in a magnetic field B. Suddenly the field is reduced
to \( \frac{B}{2} \). The radius of the circular path is
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to \( \frac{B}{2} \). The radius of the circular path is
Solution:
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Verified Answer
The correct answer is:
\( 2 r \)
The radius of the circular path followed by an electron in uniform magnetic field is given by
\[
\begin{array}{l}
r=\frac{m v}{q B} \\
\Rightarrow r \propto \frac{1}{B}
\end{array}
\]
\[
\begin{array}{l}
r=\frac{m v}{q B} \\
\Rightarrow r \propto \frac{1}{B}
\end{array}
\]
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