A proton of mass $\mathrm{m}$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the proton will move in a circular path of radius 

In a cyclotron, for a given magnet, the radius of the semicircle traced by the positive ion is directly proportional to ($v=\mathrm{velocity} \mathrm{of} \mathrm{positive} \mathrm{ion}$)

An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$ . Suddenly the field is reduced to$\frac{B}{2}$. The radius of the circular path now becomes, 

In the cyclotron, as the radius of the circular path of the charged particle increase, $\omega =$angular velocity and, $v=$linear velocity) 

In a cyclotron, the angular frequency of a charged particle is independent of

Two particles $X$ and $Y$ having equal charges after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii  $r_1$  and $r$ respectively. The ratio of the mass of $X$ to that of $Y$ is, 

A charge $\text{'}q\text{'}$ is accelerated through a potential difference,$V$ It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $\text{'}r\text{'}$. The potential difference required to move it in a circle of radius $2r$ is,

A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is $\mathrm{R}$. When energy of particle is doubled, then new radius will be 

An electron having mass $9.1\times {10}^{-31} \mathrm{kg}$, charge $1.6\times {10}^{-19} \mathrm{C}$ and moving with the velocity of ${10}^6 \mathrm{m} {\mathrm{s}}^{-1}$ enters a region where a magnetic field exists. If it describes a circle of radius $0.2 \mathrm{m}$ then the intensity of the magnetic field must be $\times {10}^{-5} \mathrm{T}$.