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

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(D) The magnetic force $F$ acting on a charged particle moving in a magnetic field is given by $F = qvB \sin 	heta$. Since the electron moves in a ci

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The magnetic force $F$ acting on a charged particle moving in a magnetic field is given by $F = qvB \sin 	heta$. Since the electron moves in a circular orbit,$	heta = 90^\circ$,so $F = evB$.$F = (1.6 	imes 10^{-19} \, C) 	imes (4 	imes 10^6 \, m/s) 	imes (2 	imes 10^{-1} \, T) = 1.28 	imes 10^{-13} \, N$.The centripetal force required for circular motion is provided by the magnetic force: $\frac{mv^2}{r} = evB$.Rearranging for the radius $r$,we get $r = \frac{mv}{eB}$.$r = \frac{(9 	imes 10^{-31} \, kg) 	imes (4 	imes 10^6 \, m/s)}{(1.6 	imes 10^{-19} \, C) 	imes (2 	imes 10^{-1} \, T)} = \frac{36 	imes 10^{-25}}{3.2 	imes 10^{-20}} = 1.125 	imes 10^{-4} \, m \approx 1.1 	imes 10^{-4} \, m$.

An electron having charge $1.6 \times 10^{-19} \, C$ and mass $9 \times 10^{-31} \, kg$ is moving with $4 \times 10^6 \, m/s$ 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 are:

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