Question

A proton of energy $8$ $eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be
A.$4$$eV$
B.$2$$eV$
C.$8$$eV$
D.$6$$eV$

Answer 1

Hint: When a charged particle moves in a uniform magnetic field, the particle experiences a magnetic force. here kinetic energy of a charged particle can be determined by substituting the equation of momentum$(mv=qBr)$of a particle in a magnetic field $B$ with the kinetic energy equation ($K=\dfrac{m{{v}^{2}}}{2}$).


Formula used:
The kinetic energy,$K$of a charged particle moving in a circular path in a uniform magnetic field,$B$can be expressed in the following way:
$K=\dfrac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}$
Here $q=$charge of the particle
$r=$ radius of the circular path
$m=$mass of the particle

Complete answer:
Here proton and alpha particles both are charged particles. When they enter with the same velocity,$v$ in a magnetic field,$B$ perpendicularly and follow the same circular path with radius $r$.
Now as we know the radius,$r$of the circular path can be written as:
$r=\dfrac{mv}{qB}$
Where $q\And B$denotes the charge and mass of the particle respectively.
Or,$mv=qBr$[since $mv$$=p$ the momentum of a charged particle]
Or,$p=qBr$ ……(i)
Also, kinetic energy$K$ can be written in terms of the velocity,$v$ and mass, $m$of the particle.
$K=\dfrac{m{{v}^{2}}}{2}$
Or,$K=\dfrac{{{m}^{2}}{{v}^{2}}}{2m}$
Or,$K=\dfrac{{{p}^{2}}}{2m}$ [since $p=mv$] …….(ii)
Substituting the value of $p$ in equation (ii),
$K=\dfrac{{{(qBr)}^{2}}}{2m}$
Or,$K=\dfrac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}$ ……(iii)
So, kinetic energy expression for proton and alpha particle can be written as:
${{K}_{p}}=\dfrac{q_{p}^{2}\times {{B}^{2}}\times {{r}^{2}}}{2{{m}_{p}}}$ …….(iv)
${{K}_{\alpha }}=\dfrac{q_{\alpha }^{2}\times {{B}^{2}}\times {{r}^{2}}}{2{{m}_{\alpha }}}$ ..……(v)
$\alpha -$particle is the nucleus of $_{2}^{4}He$an atom. So, the mass of $\alpha -$particles (${{m}_{\alpha }}$) is four times the mass of the proton (${{m}_{p}}$).
i.e,${{m}_{\alpha }}=4{{m}_{p}}$ and charge of $\alpha -$particles (${{q}_{\alpha }}$) is two times of charge of the proton (${{q}_{p}}$) i.e, ${{q}_{\alpha }}=2{{q}_{p}}$.
Putting these values in equation (v),
${{K}_{\alpha }}=\dfrac{{{(2{{q}_{p}})}^{2}}\times {{B}^{2}}\times {{r}^{2}}}{2\times 4{{m}_{p}}}=\dfrac{q_{p}^{2}\times {{B}^{2}}\times {{r}^{2}}}{2\times {{m}_{p}}}$
Or,${{K}_{\alpha }}={{K}_{p}}$ [since ${{K}_{p}}=\dfrac{q_{p}^{2}\times {{B}^{2}}\times {{r}^{2}}}{2{{m}_{p}}}$ ]
Given the energy of the proton, ${{K}_{p}}=8$$eV$
$\therefore {{K}_{\alpha }}=8$$eV$
Therefore the energy of an alpha particle in the same magnetic field will be $8$$eV$.

Thus, option (C) is correct.


Note:When a particle carrying charge enters the magnetic field and then it acts right angles to the magnetic field but when a positively charged particle enters the electric field then it moves in the direction of the electric field and a negatively charged particle moves to the opposite direction of the electric field.