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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field:

$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

$A$ proton of energy $200\, MeV$ enters a magnetic field of $5\, T$. If the direction of the field is from south to north and the motion is upward,the force acting on it will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module