$A$ charged particle moving in a magnetic field $B$ has velocity components both along $B$ and perpendicular to $B$. The path of the charged particle will be:

Two particles $A$ and $B$ have equal charges but different masses $M_{A}$ and $M_{B}$. After being accelerated through the same potential difference,they enter a region of uniform magnetic field and describe paths of radii $R_{A}$ and $R_{B}$ respectively. Then $M_{A} : M_{B}$ is

In the $xy$-plane,the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. $A$ positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$,as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$,the average speed of the particle,in $\text{m s}^{-1}$,along the $x$-axis in the time interval $T$ is. . . . . .

$A$ proton and an electron both moving with the same velocity $v$ enter into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that

In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

$A$ 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: