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Motion in a magnetic field
Appears in
Concepts tested here
- Radius in terms of kinetic energy 2
All Questions
2015 AIPMT-II 1 question
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A proton and an alpha particle both enter a region of uniform magnetic field, B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$
$R_\alpha = R_p$: $\frac{2(4m_p)K_\alpha}{(2e)^2B^2} = \frac{2m_pK_p}{e^2B^2}$
$K_\alpha = K_p = 1$ MeV
2012 AIPMT-MAINS 1 question
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A proton carrying 1 MeV kinetic energy is moving in a circular path of radius 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?$R = \frac{mv}{qB} = \frac{\sqrt{2mE}}{qB}$
For equal radii: $\frac{\sqrt{2m_pE_p}}{q_pB} = \frac{\sqrt{2m_\alpha E_\alpha}}{q_\alpha B}$
$E_\alpha = \left(\frac{q_\alpha}{q_p}\right)^2\left(\frac{m_p}{m_\alpha}\right)E_p$, with $\frac{m_p}{m_\alpha} = \frac{1}{4}$ and $\frac{q_\alpha}{q_p} = 2$
$E_\alpha = (2)^2\times\frac{1}{4}\times E_p = E_p$
$E_\alpha = 1$ MeV