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(B) The radius of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mE_K}}{qB}$.Rearranging for kine

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$4$

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(B) The radius of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mE_K}}{qB}$.Rearranging for kinetic energy $E_K$,we get $E_K = \frac{q^2 B^2 r^2}{2m}$.Since the magnetic field $B$ is the same,$E_K \propto \frac{q^2 r^2}{m}$.For the $\alpha$-particle: $q_{\alpha} = 2e$,$m_{\alpha} = 4m_p$,and radius is $r$.For the proton: $q_p = e$,$m_p = m_p$,and radius is $2r$.Taking the ratio: $\frac{(E_K)_p}{(E_K)_{\alpha}} = \frac{q_p^2 r_p^2}{m_p} 	imes \frac{m_{\alpha}}{q_{\alpha}^2 r_{\alpha}^2}$.Substituting the values: $\frac{(E_K)_p}{1 \, MeV} = \frac{e^2 (2r)^2}{m_p} 	imes \frac{4m_p}{(2e)^2 r^2}$.$\frac{(E_K)_p}{1 \, MeV} = \frac{4e^2 r^2}{m_p} 	imes \frac{4m_p}{4e^2 r^2} = 4$.Therefore,$(E_K)_p = 4 \, MeV$.

An $\alpha$-particle of $1 \, MeV$ energy moves on a circular path in a uniform magnetic field. Then,the kinetic energy of a proton in the same magnetic field for a circular path of double radius is ...... $MeV$.

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