A charged particle of 2,mu,C accelerated by a | vedclass

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(B) The radius of the path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since the particle is accelerated by a potential

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$144$

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(B) The radius of the path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since the particle is accelerated by a potential difference $V$,its kinetic energy is $K = qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.Substituting $v$ in the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Squaring both sides: $r^2 = \frac{2mV}{B^2 q} \implies m = \frac{r^2 q B^2}{2V}$.Given: $q = 2 	imes 10^{-6}\,C$,$V = 100\,V$,$B = 4 	imes 10^{-3}\,T$,$r = 3 	imes 10^{-2}\,m$.Substituting the values: $m = \frac{(3 	imes 10^{-2})^2 	imes (2 	imes 10^{-6}) 	imes (4 	imes 10^{-3})^2}{2 	imes 100}$.$m = \frac{(9 	imes 10^{-4}) 	imes (2 	imes 10^{-6}) 	imes (16 	imes 10^{-6})}{200} = \frac{288 	imes 10^{-16}}{200} = 1.44 	imes 10^{-16}\,kg = 144 	imes 10^{-18}\,kg$.

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