[QUE/EM-01011] --- EM-PROBLEM

An alpha particle travels in a circular path of radius $0.45$m in a magnetic field with $B=1.2$ w/m$^2$. Calculate (i) its speed (ii) its period of revolution, and (iii) its kinetic energy. Mass of proton particle = \(1.67\times 10^{-27}\)kg \(\approx 4\times M_p= 4\times938.27\) MeV.

Solution :

• [(i)] the magnetic force \(eBv\) must be equal to the mass times acceleration. Therefore \begin{equation*} Bev = \frac{Mv^2}{R}, \end{equation*} where \(R\) is the radius of the circular orbit. Hence \begin{equation*} v= \frac{eBR}{M} = \frac{2\times1.6 \times10^{-19}\times 1.2 \times 0.45}{4\times 1.67\times 10^{-27}}\approx 2.7 \times10^7 \text{m/s}. \end{equation*}
• [(ii)] The time period is \begin{equation*} T = \frac{2\pi R}{v} = \frac{2\times3.14\times 0.45}{2.7\times10^7} \approx10^{-7} \text{ s}. \end{equation*}
• [(iii)] The kinetic energy is given by \begin{eqnarray}\nonumber \text{K.E.} &=& \frac{1}{2} M v^2= \frac{1}{2}\times (4\times 1.67 \times 10^{-27}) \times \big(2.7\times10^7\big)^2 \\\nonumber &=& 3.26\times 7.29 \times 10^{-13} \approx 23.7 \times 10^{-13} \text{J}. \end{eqnarray}