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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since the particle is accelerated through a potential

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$R/\sqrt{2}$

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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since the particle is accelerated through a potential difference $V$,its kinetic energy is $K = qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.Substituting $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.For a proton,$q_p = e$ and $m_p = m$. Thus,$r_p = \frac{1}{B} \sqrt{\frac{2mV}{e}}$.For a deuteron,$q_d = e$ and $m_d = 2m$. Thus,$r_d = \frac{1}{B} \sqrt{\frac{2(2m)V}{e}} = \sqrt{2} 	imes \frac{1}{B} \sqrt{\frac{2mV}{e}} = \sqrt{2} r_p$.Given $r_d = R$,we have $R = \sqrt{2} r_p$,which implies $r_p = \frac{R}{\sqrt{2}}$.

$A$ proton and a deuteron are both accelerated through the same potential difference and enter a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$,assuming the neutron and proton masses are nearly equal,the radius of the proton's path will be

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