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(D) The kinetic energy of the electron accelerated by potential $V$ is given by $K = eV = \frac{p^2}{2m}$.Thus,the momentum $p = \sqrt{2meV}$.The radi

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$7.5 	imes 10^{-4} \, m$

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(D) The kinetic energy of the electron accelerated by potential $V$ is given by $K = eV = \frac{p^2}{2m}$.Thus,the momentum $p = \sqrt{2meV}$.The radius of the circular path in a magnetic field $B$ is $R = \frac{p}{eB} = \frac{\sqrt{2meV}}{eB} = \frac{1}{B} \sqrt{\frac{2mV}{e}}$.Given: $V = 500 \, V$,$B = 100 \, mT = 0.1 \, T$,$m = 9.1 	imes 10^{-31} \, kg$,$e = 1.6 	imes 10^{-19} \, C$.Substituting the values:$R = \frac{1}{0.1} \sqrt{\frac{2 	imes 9.1 	imes 10^{-31} 	imes 500}{1.6 	imes 10^{-19}}}$$R = 10 	imes \sqrt{\frac{910 	imes 10^{-31}}{1.6 	imes 10^{-19}}} = 10 	imes \sqrt{568.75 	imes 10^{-12}}$$R = 10 	imes 23.85 	imes 10^{-6} \approx 2.38 	imes 10^{-4} \, m$.Wait,re-calculating: $p = \sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 500 	imes 1.6 	imes 10^{-19}} = \sqrt{145600 	imes 10^{-50}} = 1.206 	imes 10^{-23} \, kg \cdot m/s$.$R = \frac{1.206 	imes 10^{-23}}{1.6 	imes 10^{-19} 	imes 0.1} = \frac{1.206 	imes 10^{-23}}{1.6 	imes 10^{-20}} = 0.753 	imes 10^{-3} \, m = 7.53 	imes 10^{-4} \, m$.

In an experiment,electrons are accelerated from rest by applying a voltage of $500 \, V$. Calculate the radius of the path if a magnetic field of $100 \, mT$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19} \, C$,Mass of the electron $= 9.1 \times 10^{-31} \, kg$]

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