$A$ charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when

An electron,moving in a uniform magnetic field of induction of intensity $\vec{B}$,has its radius directly proportional to

$A$ 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:

$A$ proton, a deuteron, and an $\alpha$-particle with the same kinetic energy enter a uniform magnetic field at a right angle to the magnetic field. The ratio of the radii of their respective circular paths is

$A$ particle of specific charge $q / m = \pi \text{ C kg}^{-1}$ is projected from the origin towards the positive $X$-axis with a velocity $10 \text{ ms}^{-1}$ in a uniform magnetic field $\vec{B} = -2 \hat{k} \text{ T}$. The velocity $\vec{v}$ of the particle after time $t = \frac{1}{12} \text{ s}$ will be (in $\text{ ms}^{-1}$):

$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$,then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)