$A$ charged particle is moving in a uniform magnetic field. It penetrates a layer of lead and thereby loses half of its kinetic energy. What happens to the radius of curvature of its path?

An electron of mass $m$ and charge $e$ is projected normally from the surface of a sphere of radius $a$ with speed $v_0$ into a uniform magnetic field $B$ perpendicular to the plane of the paper. The centre of the sphere is at a distance $b$ from a wall. If the electron strikes the wall symmetrically with respect to the $x$-axis,what should be the magnetic field $B$ such that the charged particle just touches the wall?

Explain: Velocity selector.

An electron is travelling horizontally towards the east. $A$ magnetic field in the vertically downward direction exerts a force on the electron along:

$A$ $5 \text{ mg}$ particle carrying a charge of $5 \times 10^{-6} \text{ C}$ is moving with a velocity of $(3\hat{i} + 2\hat{k}) \times 10^{-2} \text{ m/s}$ in a region having a magnetic field $\vec{B} = 0.1\hat{k} \text{ Wb/m}^2$. It moves a distance of $\alpha \text{ m}$ along the $\hat{k}$ direction when it completes $5$ revolutions. The value of $\alpha$ is . . . . . . .

In a mass spectrometer used for measuring the masses of ions,the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant,the ratio $\left(\frac{\text{charge on the ion}}{\text{mass of the ion}}\right)$ will be proportional to