A charge $Q$ of mass $M$ moving in a straight line is accelerated by a potential difference $V$. It enters a uniform magnetic field $B$ perpendicular to its path. Deduce an expression in terms of $V$ for the radius of the circular path in which it travels?
Answer
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Hint We will find the kinetic energy increase of the particle considering it starts initially from rest. Therefore, the energy increase of the particle after the accelerated motion is given by $QV = \dfrac{1}{2}M{v^2}$, from which we will derive the velocity. This velocity will be substituted in the general equation of the mass spectrometer, i.e. $BQv = \dfrac{{M{v^2}}}{r}$.
Complete step by step answer
The initial energy of the charge $Q$ of mass $M$ moving in a straight line by potential difference $V$ is given by $QV$.
This energy is stored as the kinetic energy of the particle which is $\dfrac{1}{2}M{v^2}$, where $v$ is the velocity at the end of the accelerated motion.
Therefore, we get
$QV = \dfrac{1}{2}M{v^2}$, or
${v^2} = \dfrac{{2QV}}{M}$.
Therefore, the velocity of the particle at the end of the accelerated motion is
$v = \sqrt {\dfrac{{2QV}}{M}} $.
Now the force on the charged particle, when it enters a magnetic field is equal to the centripetal force since the magnetic force acting on the particle is perpendicular to the direction of motion of the particle.
$\therefore $ the required relationship between the magnetic force and the centripetal force is,
$BQv = \dfrac{{M{v^2}}}{r}$
$ \Rightarrow r = \dfrac{{Mv}}{{BQ}}$
Substituting the value of the velocity from the above equation, we get
$r = \dfrac{M}{{BQ}}\sqrt {\dfrac{{2QV}}{M}} $
$ \Rightarrow r = \dfrac{1}{B}\sqrt {\dfrac{{2MV}}{Q}} $
which is the required value of the radius of the circular path.
We can clearly see that the radius of curvature not only depends on the magnetic field, but also on the initial voltage difference through which it was accelerated. It also depends on its own mass and charge.
Note The mass of the charged particle is the rest mass of the particle with charge $Q$ moving in a straight line and accelerated by the potential difference $V$. This is the general equation of a charged particle moving in a magnetic field in a mass spectrometer.
Complete step by step answer
The initial energy of the charge $Q$ of mass $M$ moving in a straight line by potential difference $V$ is given by $QV$.
This energy is stored as the kinetic energy of the particle which is $\dfrac{1}{2}M{v^2}$, where $v$ is the velocity at the end of the accelerated motion.
Therefore, we get
$QV = \dfrac{1}{2}M{v^2}$, or
${v^2} = \dfrac{{2QV}}{M}$.
Therefore, the velocity of the particle at the end of the accelerated motion is
$v = \sqrt {\dfrac{{2QV}}{M}} $.
Now the force on the charged particle, when it enters a magnetic field is equal to the centripetal force since the magnetic force acting on the particle is perpendicular to the direction of motion of the particle.
$\therefore $ the required relationship between the magnetic force and the centripetal force is,
$BQv = \dfrac{{M{v^2}}}{r}$
$ \Rightarrow r = \dfrac{{Mv}}{{BQ}}$
Substituting the value of the velocity from the above equation, we get
$r = \dfrac{M}{{BQ}}\sqrt {\dfrac{{2QV}}{M}} $
$ \Rightarrow r = \dfrac{1}{B}\sqrt {\dfrac{{2MV}}{Q}} $
which is the required value of the radius of the circular path.
We can clearly see that the radius of curvature not only depends on the magnetic field, but also on the initial voltage difference through which it was accelerated. It also depends on its own mass and charge.
Note The mass of the charged particle is the rest mass of the particle with charge $Q$ moving in a straight line and accelerated by the potential difference $V$. This is the general equation of a charged particle moving in a magnetic field in a mass spectrometer.
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