$A$ proton of mass $m$ moving with a speed $v$ ($v << c$, where $c$ is the velocity of light in vacuum) completes a circular orbit in time $T$ in a uniform magnetic field. If the speed of the proton is increased to $\sqrt{2}v$, what will be the time needed to complete the circular orbit?

$A$ charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a

What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .

$A$ stream of electrons is projected horizontally to the right. $A$ straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right,then what will be the effect on the electron stream?

The figure shows a region of length $l$ with a uniform magnetic field of $0.3 \, T$ in it. $A$ proton enters the region with a velocity of $4 \times 10^{5} \, m/s$,making an angle of $60^{\circ}$ with the field. If the proton completes $10$ revolutions by the time it crosses the region shown,$l$ is close to....... $m$ (mass of proton $= 1.67 \times 10^{-27} \, kg$,charge of the proton $= 1.6 \times 10^{-19} \, C$)