JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 13)
A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is :
1 : 4
4 : 1
2 : 1
1 : 2
Explanation
$R=\frac{m v}{q B}$
$\frac{\mathrm{R}_\alpha}{\mathrm{R}_{\mathrm{P}}}=\frac{\mathrm{M}_\alpha}{\mathrm{M}_{\mathrm{P}}} \times \frac{\mathrm{q}_{\mathrm{P}}}{\mathrm{q}_\alpha}$
$\frac{\mathrm{R}_\alpha}{\mathrm{R}_{\mathrm{P}}}=\frac{4}{1} \times \frac{1}{2}=2$
$\frac{\mathrm{R}_\alpha}{\mathrm{R}_{\mathrm{P}}}=\frac{\mathrm{M}_\alpha}{\mathrm{M}_{\mathrm{P}}} \times \frac{\mathrm{q}_{\mathrm{P}}}{\mathrm{q}_\alpha}$
$\frac{\mathrm{R}_\alpha}{\mathrm{R}_{\mathrm{P}}}=\frac{4}{1} \times \frac{1}{2}=2$
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