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An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbit of radii ${r_e},{r_p},{r_\alpha }$ respectively in a uniform magnetic field $B$. The relation between ${r_e},{r_p},{r_\alpha }$ is ?
A) ${r_e} < {r_p} < {r_\alpha }$
B) ${r_e} < {r_\alpha } < {r_p}$
C) ${r_e} > {r_p} = {r_\alpha }$
D) ${r_e} < {r_p} = {r_\alpha }$


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Last updated date: 17th Jun 2024
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Answer
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Hint: The radius of the circular path through which a particle rotates in a magnetic is proportional to the mass of the particle rotates at that circular path. Even if an electron, proton and alpha particle have negligible mass, the mass of those particles are not equal.

Complete step by step answer:
 Let’s discuss the forces which are acting on particles which travel through a circular path in a magnetic field.
As the particles are moving in a circular path, within the magnetic field, there will be centripetal force acting on the revolving particles in the circular path. And this centripetal force will be equal to the magnetic force acting on the particle caused by the magnetic through which those particles are travelling.
The centripetal force acting on the particle is given by the formula:
Centripetal force, ${F_c} = \dfrac{{m{v^2}}}{r}$
Where, $m$ = mass of the particle
$v$ = velocity of the particle
$r$ = radius of the circular path through which the particle rotates.
The magnetic force acting on the particle is given by the formula:
Magnetic force, ${F_m} = qvB$
Where, $q$ = charge of moving particle
$v$ = velocity of the particle
$B$ = magnetic field
As we have discussed earlier, centripetal force is equal to the magnetic force:
$ \Rightarrow {F_c} = {F_m}$
$ \Rightarrow \dfrac{{m{v^2}}}{r} = qvB$
\[ \Rightarrow \dfrac{{mv}}{r} = qB\]
\[ \Rightarrow v = \dfrac{{qBr}}{m}\]
We know the kinetic energy is given by the formula,
Kinetic energy, $E = \dfrac{{m{v^2}}}{2}$
Substituting the value of $v$ in this formula, we get,
$ \Rightarrow E = \dfrac{m}{2}\dfrac{{{{(qBr)}^2}}}{{{m^2}}}$
$ \Rightarrow E = \dfrac{{{{(qBr)}^2}}}{{2m}}$
From this equation we can find the value of the radius
$ \Rightarrow {r^2} = \dfrac{{2Em}}{{{{(qB)}^2}}}$
$ \Rightarrow r = \sqrt {\dfrac{{2Em}}{{{{(qB)}^2}}}} $ $ \Rightarrow r = \dfrac{{\sqrt {2Em} }}{{qB}}$

Let’s calculate the value of radius for different particles.
For electron, ${r_e} = \dfrac{{\sqrt {2k{m_e}} }}{{eB}}$
For proton ${r_p} = \dfrac{{\sqrt {2k{m_p}} }}{{eB}}$
For alpha particle, ${r_\alpha } = \dfrac{{\sqrt {2k{m_\alpha }} }}{{{q_\alpha }B}} = \dfrac{{\sqrt {2k4{m_p}} }}{{2eB}} = \dfrac{{\sqrt {2k{m_p}} }}{{eB}}$
As ${m_e} < {m_p}$
So we get, Final answer is Option (D)

Note: Centripetal force is a force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous centre of curvature of the path. The tension force in the string of a swinging tethered ball and the gravitational force keeping a satellite in orbit are both examples of centripetal forces.