An electron is accelerated by a potential dif | vedclass

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(B) The kinetic energy $K$ gained by an electron accelerated through a potential difference $V$ is given by $K = qV$.The radius $r$ of the circular pa

Vedclass · Accepted Answer

$36.7\, cm$

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(B) The kinetic energy $K$ gained by an electron accelerated through a potential difference $V$ is given by $K = qV$.The radius $r$ of the circular path of a charged particle in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB}$.Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2qV}{m}}$.Substituting $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Given values: $V = 12000\, V$,$B = 10^{-3}\, T$,$m = 9 	imes 10^{-31}\, kg$,$q = 1.6 	imes 10^{-19}\, C$.$r = \frac{1}{10^{-3}} \sqrt{\frac{2 	imes 9 	imes 10^{-31} 	imes 12000}{1.6 	imes 10^{-19}}} = 10^3 \sqrt{\frac{216 	imes 10^{-28}}{1.6 	imes 10^{-19}}} = 10^3 \sqrt{135 	imes 10^{-9}} = 10^3 \sqrt{1.35 	imes 10^{-7}} \approx 0.367\, m$.Therefore,$r = 36.7\, cm$.

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