An electron having charge 1.6 times 10^{-19} | vedclass

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(D) The magnetic force acting on a moving charge is given by $F = qvB \sin(	heta)$. Since the electron moves in a circular orbit,the velocity is perp

Vedclass · Accepted Answer

$1.28 	imes 10^{-13} \ N, 1.1 	imes 10^{-4} \ m$

Vedclass · Answer

(D) The magnetic force acting on a moving charge is given by $F = qvB \sin(	heta)$. Since the electron moves in a circular orbit,the velocity is perpendicular to the magnetic field,so $	heta = 90^\circ$ and $\sin(90^\circ) = 1$.$F = (1.6 	imes 10^{-19} \ C) 	imes (4 	imes 10^6 \ m/s) 	imes (2 	imes 10^{-1} \ T) = 1.28 	imes 10^{-13} \ N$.The magnetic force provides the necessary centripetal force for circular motion,so $F = \frac{mv^2}{r}$.Rearranging for radius $r$: $r = \frac{mv}{qB} = \frac{(9 	imes 10^{-31} \ kg) 	imes (4 	imes 10^6 \ m/s)}{1.6 	imes 10^{-19} \ C 	imes 2 	imes 10^{-1} \ T}$.$r = \frac{36 	imes 10^{-25}}{3.2 	imes 10^{-20}} = 11.25 	imes 10^{-5} \ m \approx 1.1 	imes 10^{-4} \ m$.

An electron having charge $1.6 \times 10^{-19} \ C$ and mass $9 \times 10^{-31} \ kg$ is moving with $4 \times 10^6 \ m \ s^{-1}$ speed in a magnetic field of $2 \times 10^{-1} \ T$ in a circular orbit. The force acting on the electron and the radius of the circular orbit is $.........$

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