In a region,steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. $A$ charged particle is released from rest in this region. The path of the particle will be a

$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .

$A$ magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region,a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. $A$ narrow beam of (single species) charged particles all accelerated through $15 \;kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{5} \;V \,m^{-1}$,make a simple guess as to what the beam contains. Why is the answer not unique?

If a proton, deuteron, and $\alpha$-particle are accelerated by the same potential difference and enter perpendicular to a magnetic field, then the ratio of their kinetic energies is:

$A$ charged particle, when entering a uniform magnetic field, moves in a helical path. If its angular velocity is $4 \pi \times 10^6 \text{ rad s}^{-1}$ and its velocity in the direction of the magnetic field is $3 \times 10^5 \text{ m s}^{-1}$, then the pitch of the helix is: (in $\text{ cm}$)

$A$ charged particle of specific charge $\alpha$ is released from the origin at time $t = 0$ with velocity $\vec{V} = V_o \hat{i} + V_o \hat{j}$ in a magnetic field $\vec{B} = B_o \hat{i}$. The coordinates of the particle at time $t = \frac{\pi}{B_o \alpha}$ are (specific charge $\alpha = q/m$):