Question

A $1.5\,m$ diameter cyclotron is used to accelerate protons to an energy of $8\,MeV$. The required magnetic field strength for this cyclotron is: $({m_p} = 1.6 \times {10^{ - 27}}\,Kg)\,N$

Answer 1

Hint: In order to this question, to calculate the magnetic field strength for the given cyclotron, we will apply the formula of Kinetic energy as kinetic energy is applied to accelerate the protons.

Formula used:
${E_k} = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}}$
Here, $q$ is the charge of the particle, $B$ is the magnetic field strength, what we have to find, $r$ is the radius of the cyclotron and $m$ is the mass of the proton.

Complete step by step answer:
To accelerate an object we have to apply force. To apply force, we need to do work. When work is done on an object, energy is transferred and the object moves with a new constant speed. The energy that is transferred is known as kinetic energy and it depends on the mass and speed achieved. As we know, ${E_k}$ is the kinetic energy which is provided energy to the proton to accelerate, i.e..
${E_k} = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}}$
$\Rightarrow 8 \times {10^6} \times 1.6 \times {10^{ - 9}}J = \dfrac{{{{(1.6 \times {{10}^{ - 19}})}^2} \times {B^2} \times {{(\dfrac{{1.5}}{2})}^2}}}{{2 \times 1.6 \times {{10}^{ - 27}}}} \\
\Rightarrow B = \sqrt {\dfrac{{8 \times {{10}^6} \times 2 \times (1.6 \times {{10}^{ - 27}})}}{{1.6 \times {{10}^{ - 19}} \times {{(\dfrac{{1.5}}{2})}^2}}}} = 5.33 \times {10^{ - 1}} \\
\therefore B = 0.533\,T \\ $
Hence, the required magnetic field strength for the given cyclotron is $0.533\,T$.

Note: A cyclotron is a device that boosts the energy of charged particles or ions. E.O Lawrence and M.S Livingston invented it in 1934 to investigate the nuclear structure. The cyclotron uses both electric and magnetic fields to boost the energy of charged particles.Cross fields are named so because both fields are perpendicular to each other.