The radius of the path of an electron moving | vedclass

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(C) The radius $R$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula: $R = \frac{mv}{qB}$.Gi

Vedclass · Accepted Answer

$30$

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(C) The radius $R$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula: $R = \frac{mv}{qB}$.Given values are:Mass of electron,$m = 9 	imes 10^{-31} \ kg$Velocity of electron,$v = 3.2 	imes 10^7 \ m/s$Charge of electron,$q = 1.6 	imes 10^{-19} \ C$Magnetic field,$B = 6 	imes 10^{-4} \ T$Substituting these values into the formula:$R = \frac{(9 	imes 10^{-31}) 	imes (3.2 	imes 10^7)}{(1.6 	imes 10^{-19}) 	imes (6 	imes 10^{-4})}$$R = \frac{28.8 	imes 10^{-24}}{9.6 	imes 10^{-23}}$$R = \frac{28.8}{9.6} 	imes 10^{-1} \ m$$R = 3 	imes 10^{-1} \ m = 0.3 \ m = 30 \ cm$.

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $6 \times 10^{-4} \ T$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \ kg$ and charge of electron is $1.6 \times 10^{-19} \ C$). (in $cm$)

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