$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field:

$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

$A$ charge $Q$ is moving through a distance $d\vec{l}$ in a magnetic field $\vec{B}$. Find the value of the work done by the magnetic field.

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?

$A$ proton of mass $1.67 \times 10^{-27} \, kg$ and charge $1.6 \times 10^{-19} \, C$ is projected with a speed of $2 \times 10^6 \, m/s$ at an angle of $60^\circ$ to the $X$-axis. If a uniform magnetic field of $0.104 \, T$ is applied along the $Y$-axis,the path of the proton is: