Q126 / 180 · KCET 2018 · Physics
A charge \( q \) is accelerated through a potential difference \( V \). It is then passed normally through
a uniform magnetic field, where it moves in a circle of radius \( \mathrm{r} \). The potential difference
required to move it in a circle of radius \( 2 r \) is
a uniform magnetic field, where it moves in a circle of radius \( \mathrm{r} \). The potential difference
required to move it in a circle of radius \( 2 r \) is
- A\( 12 \mathrm{~V} \)
- Correct answerB4V
- C\( 1 \mathrm{~V} \)
- D\( 3 \mathrm{~V} \)
Solution
Given, charge $\mathrm{q}$ is accelerated through a potential difference $\mathrm{V}$. Therefore,
$\frac{1}{2} m v^{2}=q V$
$\Rightarrow m v^{2}=2 q V$
$\Rightarrow(m v)^{2}=2 q m V$
$\Rightarrow m v=\sqrt{2 m q V}$
The radius of circular path followed by the charge is
$r=\frac{m v}{q B}=\frac{\sqrt{2 m q V}}{q B}$
$\Rightarrow r \propto \sqrt{V}$
Thus, the potential difference required to move the charge in circle of radius $2 r$ is $4 \mathrm{~V}$
Hence, the correct answer is option: 'B'.