The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

$A$ 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:

$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field 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 $= 9.0 \times 10^{-31} \, kg$)

$A$ proton (or charged particle) moving with velocity $v$ is acted upon by an electric field $E$ and a magnetic field $B$. The proton will move undeflected if