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  • I have a doubt, kindly clarify. - Magnetic Effects of Current and Magnetism - NEET

A proton carrying 1MeV kinetic energy is moving in a circular path of redius R in uniform magnetic field. What should be the energy of an \alpha-particle to describe a circle of same radius in the same field?

  • Option 1)

    2MeV

  • Option 2)

    1MeV

  • Option 3)

    0.5MeV

  • Option 4)

    4MeV

 

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}

-

 

 r=\frac{\sqrt{2mE}}{qB}\frac{\gamma _{\alpha }}{\gamma _{p}}=\frac{\sqrt{2m\alpha E\alpha }}{q_{\alpha B.\frac{\sqrt{2mpEP}}{qpB}}}=\frac{q_{p}}{q\alpha }.\sqrt{\frac{m_{\alpha E_{\alpha }}}{E_{p}mp}}

\gamma _{p}=\gamma _{\alpha }=R;m_{\alpha }=4mp\:and\:q_{\alpha }=2qp\\ E_{\alpha }=E_{p}=1MeV


Option 1)

2MeV

This solution is incorrect 

Option 2)

1MeV

This solution is correct 

Option 3)

0.5MeV

This solution is incorrect 

Option 4)

4MeV

This solution is incorrect 

Posted by

prateek

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