A proton of energy $200\, MeV$ enters the magnetic field of $5\, T$. If direction of field is from south to north and motion is upward, the force acting on it will be

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

A electron $(q = 1.6 \times 10^{-19}\, C)$ is moving at right angle to the uniform magnetic field $3.534 \times 10^{-5}\, T$. The time taken by the electron to complete a circular orbit is......$µs$

If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let $r_p, r_e$ and $r_{He}$ be their respective radii, then

A charge particle moving in magnetic field $B$, has the components of velocity along $B$ as well as perpendicular to $B$. The path of the charge particle will be