
Show that for a given positive ion in a cyclotron
(a) The radius of their circular path inside it is proportional to the velocity.
(b) The time spent inside the cyclotron is dependent on radius and speed.
Answer
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Hint: In order to solve this question, we should know about the working principle of cyclotron. A cyclotron is a device in which atomic particles are accelerated by changing electric field in the presence of magnetic field inside the device, this produces a centripetal force on the particle and is balanced by Lorentz magnetic force on it. We will use this concept to solve both parts of a given problem.
Formula used:
If $B$ is the magnetic field, $v$ is the velocity of the particle $q$ charge of the particle and $r$ is the radius of the circular path of the particle inside the cyclotron having mass m of the particle then Lorentz magnetic force acting on particle is ${F_B} = Bqv$ and centripetal force acting on particle is ${F_C} = \dfrac{{m{v^2}}}{r}$
Complete step by step answer:
By principle of cyclotron we know, the centripetal force acting on positive ion is balanced by the Lorentz magnetic force acting on it so,
We have centripetal force, ${F_C} = \dfrac{{m{v^2}}}{r}$
Magnetic Lorentz force, ${F_B} = Bqv$
Equation both forces we get,
$\dfrac{{m{v^2}}}{r} = Bqv$
$\Rightarrow \dfrac{{mv}}{r} = Bq$
$\Rightarrow rBq = mv$
From above relation we can see that,
(a) Radius is directly proportional to the velocity of the positive ion i.e, $r \propto v$.
(b) Now, if $r$ is the radius of circular path and $v$ is the velocity of the positive ion then, total circular path covered by positive ion is $2\pi r$ then, using relation of $\text{speed} = \dfrac{\text{Distance}}{\text{time}}$ we get,
$v = \dfrac{{2\pi r}}{T}$
$\therefore T = \dfrac{{2\pi r}}{v}$
Hence, from the above relation, we can see that time spent inside a cyclotron depends upon the radius $r$ and velocity $v$ of the positive ion.
Note: It should be remembered that, frequency of the atomic particle inside the cyclotron is always independent of its velocity and radius of circular path as it’s calculated by the relation $f = \dfrac{{qB}}{{2\pi m}}$ and cyclotron is widely used in particle physics to accelerate atomic particles at high velocities and then study the nature of these particles.
Formula used:
If $B$ is the magnetic field, $v$ is the velocity of the particle $q$ charge of the particle and $r$ is the radius of the circular path of the particle inside the cyclotron having mass m of the particle then Lorentz magnetic force acting on particle is ${F_B} = Bqv$ and centripetal force acting on particle is ${F_C} = \dfrac{{m{v^2}}}{r}$
Complete step by step answer:
By principle of cyclotron we know, the centripetal force acting on positive ion is balanced by the Lorentz magnetic force acting on it so,
We have centripetal force, ${F_C} = \dfrac{{m{v^2}}}{r}$
Magnetic Lorentz force, ${F_B} = Bqv$
Equation both forces we get,
$\dfrac{{m{v^2}}}{r} = Bqv$
$\Rightarrow \dfrac{{mv}}{r} = Bq$
$\Rightarrow rBq = mv$
From above relation we can see that,
(a) Radius is directly proportional to the velocity of the positive ion i.e, $r \propto v$.
(b) Now, if $r$ is the radius of circular path and $v$ is the velocity of the positive ion then, total circular path covered by positive ion is $2\pi r$ then, using relation of $\text{speed} = \dfrac{\text{Distance}}{\text{time}}$ we get,
$v = \dfrac{{2\pi r}}{T}$
$\therefore T = \dfrac{{2\pi r}}{v}$
Hence, from the above relation, we can see that time spent inside a cyclotron depends upon the radius $r$ and velocity $v$ of the positive ion.
Note: It should be remembered that, frequency of the atomic particle inside the cyclotron is always independent of its velocity and radius of circular path as it’s calculated by the relation $f = \dfrac{{qB}}{{2\pi m}}$ and cyclotron is widely used in particle physics to accelerate atomic particles at high velocities and then study the nature of these particles.
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