Amity University Noida B.Tech Admissions 2025
ApplyAmong Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
14 Questions around this concept.
Direction: In the following question, a statement of Assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
Assertion: Cyclotron is an accelerating device.
Reason: Neutrons can not be accelerated by the cyclotron.
In a cyclotron, a charged particle
Cyclotron
Cyclotron is a device used to accelerated positively charged particles (like, a-particles, deutrons etc.) to acquire enough energy to carry out nuclear disintegration etc. It is based on the fact that the electric field accelerates a charged particle and the magnetic field keeps it revolving in circular orbits of constant frequency.
It consists of two hollow D -shaped metallic chambers $D_1$ and $D_2$ called dees. The two dees are placed horizontally with a small gap separating them. The dees are connected to the source of high frequency electric field. The dees are enclosed in a metal box containing a gas at a Iow pressure of the order of $10^{-3} \mathrm{~mm}$ mercury. The whole apparatus is placed between the two poles of a strong electromagnet NS as shown in fig. The magnetic field acts perpendicular to the plane of the dees.|
Radius of the path travelled by the particle can be given as:
$
r=\frac{m V}{q B}
$
where V is the velocity, q is the charge and B is magnitude of magnetic field applied.
Cyclotron frequency : Time taken by ion to describe a semicircular path is given by
$
t=\frac{\pi r}{v}=\frac{\pi m}{q B}
$
If $\mathrm{T}=$ time period of oscillating electric field then,
$
T=2 t=\frac{2 \pi m}{q B}
$
therefore the cyclotron frequency
$
\nu=\frac{1}{T}=\frac{B q}{2 \pi m}
$
Maximum energy of particle : Maximum energy gained by the charged particle
$
E_{\max }=\left(\frac{q^2 B^2}{2 m}\right) r_0^2
$
where $r_0=$ maximum radius of the circular path followed by the positive ion.
"Stay in the loop. Receive exam news, study resources, and expert advice!"