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(A) When a charged particle of mass $m$ and charge $q$ enters a uniform magnetic field $B$ with a velocity $v$ perpendicular to the field,it undergoes

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same for all electrons

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(A) When a charged particle of mass $m$ and charge $q$ enters a uniform magnetic field $B$ with a velocity $v$ perpendicular to the field,it undergoes circular motion.The magnetic Lorentz force provides the necessary centripetal force: $qvB = \frac{mv^2}{r}$.From this,the radius of the circular path is $r = \frac{mv}{qB}$.The time period $T$ is the time taken to complete one full revolution,given by $T = \frac{2\pi r}{v}$.Substituting the value of $r$: $T = \frac{2\pi (mv/qB)}{v} = \frac{2\pi m}{qB}$.Since $m$,$q$,and $B$ are constants for all electrons in this scenario,the time period $T$ is independent of the speed $v$.Therefore,the time periods of rotation will be the same for all electrons.

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be:

Similar Questions

$A$ charged particle with charge $q$ enters a region of constant,uniform,and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B}$,and comes out without any change in the magnitude or direction of $\vec{v}$. Then:

$A$ particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$. If an electric field $\vec{E} = E_{0} \hat{i}$ and a magnetic field $\vec{B} = B_{0} \hat{i}$ act on the particle,its speed will double after a time:

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.
Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.

$A$ charged particle is moving along a magnetic field line. What is the magnetic force acting on the particle? $(\sin 0^{\circ}=0, \sin \frac{\pi}{2}=1)$

An electron with energy $0.1 \, keV$ moves at a right angle to the Earth's magnetic field of $1 \times 10^{-4} \, Wb/m^2$. The frequency of revolution of the electron will be. (Take mass of electron $= 9.0 \times 10^{-31} \, kg$)

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