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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 8 of 27
Marks: +1, -0
An α\alpha-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field B→\overset{\rightarrow}{B}, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them.
Solution:  
Formula
Calculation of ratio of radii
   radius   r=  mvqB=  2mkqB\; \text{ radius } \; r = \; \frac{m v}{q B} = \; \frac{\sqrt{2 m k}}{q B}
Ka=K   proton   K_a = K_{\; \text{ proton } \;}
Ma=4mpM_a = 4 m_p
qa=2qpq_a = 2 q_p
  rαrp=  2mαKqαB2mpKqpB\; \frac{r_\alpha}{r_p} = \; \frac{ \frac{\sqrt{2 m_\alpha K}}{q_\alpha B} }{ \frac{\sqrt{2 m_p K}}{q_p B} }
=mαmp×qpqα= \sqrt{ \frac{m_\alpha}{m_p} } \times \frac{q_p}{q_\alpha}
=4×12=1= \sqrt{4} \times \frac{1}{2} = 1
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