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 $= 9.0 \times 10^{-31} \, kg$)

Two long parallel conductors $S_{1}$ and $S_{2}$ are separated by a distance $10 \, cm$ and carry currents of $4 \, A$ and $2 \, A$ respectively. The conductors are placed along the $x$-axis in the $X-Y$ plane. There is a point $P$ located between the conductors (as shown in the figure). $A$ charged particle of $3 \pi \, C$ is passing through the point $P$ with velocity $\overrightarrow{v} = (2 \hat{i} + 3 \hat{j}) \, m/s$; where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$ axes respectively. The force acting on the charged particle is $4 \pi \times 10^{-5} (-x \hat{i} + 2 \hat{j}) \, N$. The value of $x$ is:

An alpha particle moving with a certain speed towards the east enters a uniform magnetic field directed vertically up. The alpha particle will then move in:

$A$ proton and an alpha-particle moving with the same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of the radii of their circular paths is

$A$ charged particle moves with velocity $\overrightarrow{v}$ in a uniform magnetic field $\overrightarrow{B}$. The magnetic force experienced by the particle is

If a particle of charge $10^{-12} \, C$ moving along the $\hat{x}$ direction with a velocity $10^5 \, m/s$ experiences a force of $10^{-10} \, N$ in the $\hat{y}$ direction due to a magnetic field,then the minimum magnetic field is: