A cyclotron is accelerating proton where the applied magnetic field is 2 T, the potential gap is...

Question:

A cyclotron is accelerating proton where the applied magnetic field is 2 T, the potential gap is 10{eq}KV {/eq} then how much turn the proton has to moving between these to acquire a kinetic energy 20{eq}meV {/eq} a) 200 b) 300 c) 150 d)100

The change in Kinetic Energy and Electric Potential Difference:

The change in kinetic energy of a charged particle when it passes through a potential difference {eq}\Delta V {/eq} is given by the equation;

{eq}\Delta (K.E.) = q \Delta V {/eq}

where, {eq}q {/eq} is the charge of the charge particle and {eq}\Delta (K.E.) {/eq} is the change in the kinetic energy.

Answer and Explanation:

Given

• The applied magnetic field is {eq}B = 2\ T {/eq}

• The potential difference is {eq}\Delta V = 10\ KV {/eq}

• The kinetic energy (K.E.) of the proton is {eq}K.E. = 20\ MeV {/eq}

The change in the kinetic energy of the proton in one revolution is {eq}2 (e\Delta V) {/eq}.

If n is the number of turns made by the proton before leaving the cyclotron then,

{eq}\begin{align} K.E._{max} &= n\times (2e\Delta V)\\ \implies 20\ MeV &= n\times (2e\times 10\ KV)\\ \implies n &=\dfrac{20\ MeV}{ 20\ KeV}\\ &= 1000\\ \end{align} {/eq}

Therefore, the number of turns made by the proton before leaving the cyclotron is 1000.

None of the options is correct.