$A$ positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will

$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes $3$ times the original, the new radius will be

An ion beam of specific charge $5 \times 10^7 \ C/kg$ enters a uniform magnetic field of $4 \times 10^{-2} \ T$ with a velocity $2 \times 10^5 \ m/s$ perpendicularly. The radius of the circular path of the ions in meters will be:

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

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

$A$ charged particle with charge $q$ enters a region of constant,uniform,and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B}$,and comes out without any change in the magnitude or direction of $\vec{v}$. Then: