# A deuteron of kinetic energy $50\,{\text{keV}}$ is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field $B$. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same path with the same $B$ isA. $25\,{\text{keV}}$B. $50\,{\text{keV}}$C. $200\,{\text{keV}}$D. $100\,{\text{keV}}$

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Hint: Use the formula for centripetal force and force due to the magnetic field acting on a particle moving in a circular orbit. The force acting on the particle due to the magnetic field is equal to the centripetal force. From this relation, derive the equation for kinetic energy of the particle and solve it for kinetic energy of proton.

Formulae used:
The centripetal force ${F_C}$ acting on an object in circular motion is
${F_C} = \dfrac{{m{v^2}}}{R}$ …… (1)
Here, $m$ is the mass of the object, $v$ is the velocity of the object and $R$ is the radius of the circular path.
The magnetic force ${F_B}$ acting on an particle is
${F_B} = qvB$ …… (2)
Here, $q$ is charged on the particle, $v$ is the velocity of the particle and $B$ is the magnetic field.
The kinetic energy $K$ of an object is
$K = \dfrac{1}{2}m{v^2}$ …… (3)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.

We have given that the kinetic energy of a deuteron is $50\,{\text{keV}}$.
${K_d} = 50\,{\text{keV}}$
The radius of the circular orbit in which the deuteron and proton are moving is $0.5\,{\text{m}}$.
$\Rightarrow R = 0.5\,{\text{m}}$
Let ${m_p}$ be the mass of the proton and ${m_d}$ be the mass of the deuteron.Let us assume that the deuteron and proton are moving in the circular orbit with the same velocity.We are asked to determine the kinetic energy of the proton.For the deuteron moving in the circular orbit, the magnetic force ${F_B}$ acting on it is equal to the centripetal force ${F_C}$ acting on the deuteron.
${F_B} = {F_C}$
Substitute for ${F_B}$ and $\dfrac{{{m_d}{v^2}}}{R}$ for ${F_C}$ in the above equation.
$q{v_d}B = \dfrac{{{m_d}{v^2}}}{R}$
$\Rightarrow {m_d}{v^2} = qvBR$

Therefore, according to equation (3), the kinetic energy ${K_d}$ of deuteron is given by
$\Rightarrow {K_d} = \dfrac{1}{2}{m_d}{v^2} = \dfrac{{qvBR}}{2}$
$\Rightarrow {K_d} = \dfrac{{qBR}}{{{m_d}v}}$ …… (4)
Similarly, the kinetic energy ${K_p}$ of proton moving in the circular orbit is
$\Rightarrow {K_p} = \dfrac{1}{2}{m_p}{v^2} = \dfrac{{qvBR}}{2}$
$\Rightarrow {K_p} = \dfrac{{qBR}}{{{m_p}v}}$ …… (5)
We know that the mass of the deuteron is twice the mass of the proton.
${m_d} = 2{m_p}$

Let us divide equation (5) by equation (4).
$\Rightarrow \dfrac{{{K_p}}}{{{K_d}}} = \dfrac{{\dfrac{{qBR}}{{{m_p}v}}}}{{\dfrac{{qBR}}{{{m_d}v}}}}$
$\Rightarrow \dfrac{{{K_p}}}{{{K_d}}} = \dfrac{{{m_d}}}{{{m_p}}}$
Substitute $2{m_p}$ for ${m_d}$ in the above equation.
$\Rightarrow \dfrac{{{K_p}}}{{{K_d}}} = \dfrac{{2{m_p}}}{{{m_p}}}$
$\Rightarrow {K_p} = 2{K_d}$
Substitute $50\,{\text{keV}}$ for ${K_d}$ in the above equation.
$\Rightarrow {K_p} = 2\left( {50\,{\text{keV}}} \right)$
$\therefore {K_p} = 100\,{\text{keV}}$
Therefore, the kinetic energy of a proton is $100\,{\text{keV}}$.

Hence, the correct option is D.

Note:The students should keep in mind that while deriving the equation for kinetic energy of the proton and deuteron, the velocity of proton and deuteron should be considered the same as both the proton and deuteron are moving in the same circular orbit and same magnetic field B. If the velocities of proton and deuteron are taken different then we will not be able to reach to the answer.