Q125 / 180 · KCET 2018 · Physics
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
to \( \frac{B}{2} \). The radius of the circular path is
- A\( \frac{r}{2} \)
- Correct answerB\( 2 r \)
- C\( \frac{r}{4} \)
- D\( 4 r \)
Solution
The radius of the circular path followed by an electron in uniform magnetic field is given by
\[ \begin{array}{l} r=\frac{m v}{q B} \\ \Rightarrow r \propto \frac{1}{B} \end{array} \]
\[ \begin{array}{l} r=\frac{m v}{q B} \\ \Rightarrow r \propto \frac{1}{B} \end{array} \]
Hence, the correct answer is option: 'B'.