A deuteron and a proton moving with equal kin | vedclass

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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv

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$2$

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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.Thus,$r = \frac{\sqrt{2mK}}{qB}$.For a deuteron $(d)$ and a proton $(p)$:Mass of deuteron $m_d = 2m_p$,charge $q_d = e$.Mass of proton $m_p = m_p$,charge $q_p = e$.Given kinetic energies are equal $(K_d = K_p = K)$.Ratio of radii: $\frac{r_d}{r_p} = \frac{\sqrt{2m_d K} / eB}{\sqrt{2m_p K} / eB} = \sqrt{\frac{m_d}{m_p}} = \sqrt{\frac{2m_p}{m_p}} = \sqrt{2}$.Comparing $\frac{r_d}{r_p} = \sqrt{2} : 1$ with $\sqrt{x} : 1$,we get $x = 2$.

$A$ deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at a right angle to the field. If $r_{d}$ and $r_{p}$ are the radii of their circular paths respectively,then the ratio $\frac{r_{d}}{r_{p}}$ will be $\sqrt{x} : 1$ where $x$ is ..........

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