$A$ charged particle is moving along a magnetic field line. What is the magnetic force acting on the particle? $(\sin 0^{\circ}=0, \sin \frac{\pi}{2}=1)$

$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be

To which of the following quantities,the radius of the circular path of a charged particle moving at right angles to a uniform magnetic field is directly proportional?

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

$A$ particle of mass $M$ and charge $q$ moves with a constant velocity $V$ in the positive $x$-direction. $A$ constant magnetic field $B$ exists in the negative $z$-direction between $x = a$ and $x = b$. Find the minimum value of $V$ such that the particle enters the region $x > b$.

In a chamber,a uniform magnetic field of $6.5 \;G \left(1 \;G = 10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. $\left(e = 1.6 \times 10^{-19} \;C, m_{e} = 9.1 \times 10^{-31} \;kg \right)$