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Q

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)
104 Views

 

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 

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