If cathode rays are projected at right angles to a magnetic field,their trajectory is

$A$ narrow electron beam passes undeviated through an electric field $E = 3 \times 10^4 \ V/m$ and an overlapping magnetic field $B = 2 \times 10^{-3} \ Wb/m^2$. If the electric field and magnetic field are mutually perpendicular,the speed of the electrons is:

$A$ particle is moving in a uniform magnetic field,then

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

$A$ particle of charge $1.0 \times 10^{-16} \ C$ moves through a uniform magnetic field $B=B_0(\hat{i}+4 \hat{j}) \ T$. The particle velocity at some instant is $v=(2 \hat{i}+4 \hat{j}) \ ms^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \ N$. The magnitude of $B_0$ is (in $T$)

$A$ proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to the direction of the magnetic field. If both the particles move around the magnetic field in circles of equal radii,the ratio of the momentum of the proton to the alpha particle $\left( \frac{P_p}{P_\alpha} \right)$ is