$A$ particle of charge $q$ and mass $m$ moving with a velocity $v$ along the $x$-axis enters the region $x > 0$ with a uniform magnetic field $B$ along the $\hat{k}$ direction. The particle will penetrate this region in the $x$-direction up to a distance $d$ equal to:

An electron (mass = $9.0 \times 10^{-31} \ kg$ and charge = $1.6 \times 10^{-19} \ C$) is moving in a circular orbit in a magnetic field of $1.0 \times 10^{-4} \ Wb/m^2$. Its period of revolution is:

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field:

When a charged particle enters a uniform magnetic field,its kinetic energy

$A$ particle with charge-to-mass ratio $\frac{q}{m} = \alpha$ is shot with a speed $v$ towards a wall at a distance $d$,moving perpendicular to the wall. The minimum value of the magnetic field $\vec{B}$ that exists in this region,perpendicular to the velocity of the particle,such that the particle does not hit the wall is: