An electron with energy 880 ,eV enters a unif | vedclass

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(C) The magnetic force provides the necessary centripetal force for circular motion: $evB = \frac{mv^2}{r}$.Rearranging for radius $r$: $r = \frac{mv}

Vedclass · Accepted Answer

$4 \,cm$

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(C) The magnetic force provides the necessary centripetal force for circular motion: $evB = \frac{mv^2}{r}$.Rearranging for radius $r$: $r = \frac{mv}{eB}$.Since kinetic energy $E = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2E}{m}}$.Substituting $v$ into the radius formula: $r = \frac{\sqrt{2mE}}{eB}$.Given: $E = 880 	imes 1.6 	imes 10^{-19} \,J$,$m = 9.1 	imes 10^{-31} \,kg$,$B = 2.5 	imes 10^{-3} \,T$,and $e = 1.6 	imes 10^{-19} \,C$.$r = \frac{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 880 	imes 1.6 	imes 10^{-19}}}{1.6 	imes 10^{-19} 	imes 2.5 	imes 10^{-3}}$.$r = \frac{\sqrt{25.6256 	imes 10^{-47}}}{4 	imes 10^{-22}} \approx \frac{5.06 	imes 10^{-23}}{4 	imes 10^{-22}} \approx 0.04 \,m = 4 \,cm$.

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be:

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