$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

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

When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is:

$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

An electron of mass $m$ and charge $q$ is moving with a speed $v$ in a circular path of radius $r$ perpendicular to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved,what will be the radius of the resulting path?