When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference…

When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $\mathrm{V}_1$, it experiences a force F . If the potential difference is changed to $\mathrm{V}_2$, the force experienced by the electron in same magnetic field is 2 F , then the ratio of potential differences $\frac{\mathrm{V}_2}{\mathrm{~V}_1}$ is
  1. $2: 1$
  2. $1: 4$
  3. $4: 1$
  4. $1: 2$

Solution

$\mathrm{K} \cdot \mathrm{E}=\frac{1}{2} \mathrm{mv}^2=\mathrm{eV} \Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
Force on electron in magnetic field is $\begin{aligned} & \mathrm{F}=\mathrm{eVB}=\mathrm{e}\left(\sqrt{\frac{2 \mathrm{eV}}{\mathrm{~m}}}\right) \cdot \mathrm{B} \Rightarrow \mathrm{~F} \propto \sqrt{\mathrm{~V}} \\ & \therefore \frac{\mathrm{~F}_2}{\mathrm{~F}_1}=\sqrt{\frac{\mathrm{V}_2}{\mathrm{~V}_1}} \Rightarrow \frac{\mathrm{~V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{F}_2}{\mathrm{~F}_1}\right)^2=\left(\frac{2 \mathrm{~F}}{\mathrm{~F}}\right)^2=4: 1 \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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