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  • Help me solve this - A proton and an - particle ( with their masses in the ratio of 1 : 4 and charges in the ratio 1 : 2 - Magnetic Effects of Current and Magnetism - JEE Main

A proton and an \alpha - particle ( with their masses in the ratio of 1 : 4 and charges in the ratio 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicualr to their velocities , the ratio of the radii r_{p} : r_{\alpha} of the circular paths described by them will be :

  • Option 1)

    1 : \sqrt{2}

  • Option 2)

     1 : 3

  • Option 3)

    1 : \sqrt{3}

  • Option 4)

    1 : 2

Answers (1)

best_answer

 

Radius of charged particle -

r=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{}2mk}{qB}=\frac{1}{B}\sqrt{}\frac{2mV}{q}

\frac{M_{p}}{M_{\alpha }}=\frac{1}{4}

\frac{Q_{p}}{Q_{\alpha }}=\frac{1}{2}

Potential Difference=V

kE=q\Delta V

r=\frac{\sqrt{2mq\Delta v}}{qB}

r\alpha \sqrt{\frac{m}{q}}

\frac{r_{p}}{r_{\alpha }}=\sqrt{\frac{m_{p}}{m_{\alpha }}}\times \sqrt{\frac{q_{\alpha }}{q_{p}}}

\frac{r_{p}}{r_{\alpha }}=\sqrt{\frac{1}{4}}\times \sqrt{2}=\frac{1}{\sqrt{2}}

 

\frac{r_{p}}{r_{\alpha }}=\frac{1}{\sqrt{2}}


Option 1)

1 : \sqrt{2}

Option 2)

 1 : 3

Option 3)

1 : \sqrt{3}

Option 4)

1 : 2

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