$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be

$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

$A$ positive charge $q$ of mass $m$ is moving along the $+x$ axis with velocity $v$. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction,crossing the $y$ axis at a distance $d$. Then:

The figure shows a region of length $l$ with a uniform magnetic field of $0.3 \, T$ in it. $A$ proton enters the region with a velocity of $4 \times 10^{5} \, m/s$,making an angle of $60^{\circ}$ with the field. If the proton completes $10$ revolutions by the time it crosses the region shown,$l$ is close to....... $m$ (mass of proton $= 1.67 \times 10^{-27} \, kg$,charge of the proton $= 1.6 \times 10^{-19} \, C$)

An electron (charge $q$ $C$) enters a magnetic field of $B$ $Wb/m^2$ with a velocity of $v$ $m/s$ in the same direction as that of the field. The force on the electron is: