If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is

The magnetic field is uniform for $y > 0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y < 0$. $A$ proton,denoted by a filled circle,leaves $y = 0$ in the $-y$-direction with some speed,as shown below. Which of the following best denotes the trajectory of the proton?

If the velocity of an electron is $(2 \hat{i} + 3 \hat{j}) \text{ m s}^{-1}$ and it enters a magnetic field of $4 \hat{k} \text{ T}$,then . . . . . . .

$A$ particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{k}$. Its velocity at $(x_0, y_0, 0)$ is $(4 \hat{i} + 3 \hat{j})$. The value of $x_0$ is:

An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $A$ lies in the same plane at a distance $2R$ from the centre. Point $A$ records the magnetic field produced by the $\alpha$ particle. If the minimum time interval between two successive times at which $A$ records zero magnetic field is $t$,find the angular speed $\omega$ in terms of $t$.

When a charged particle enters a uniform magnetic field,its kinetic energy