An electron moves at right angles to a magnetic field of $1.5 \times 10^{-2} \, T$ with a speed of $6 \times 10^7 \, m/s$. If the specific charge of the electron is $1.7 \times 10^{11} \, C/kg$,the radius of the circular path will be ...... $cm$.

$A$ small block of mass $20 \,g$ and charge $4 \,mC$ is released on a long smooth inclined plane of inclination angle $45^{\circ}$. $A$ uniform horizontal magnetic field of $1 \,T$ is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is (in $\,s$)

$A$ 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?

An electron is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. The angle $\theta$ between $\vec{v}$ and $\vec{B}$ lies between $0^\circ$ and $\frac{\pi}{2}$. Its velocity vector $\vec{v}$ returns to its initial value in a time interval of:

Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

$A$ proton accelerated by a potential difference of $500 \ kV$ flies through a uniform transverse magnetic field of $0.1 \ T$. The field is spread over a region of $1.0 \ cm$ thickness. The angle through which the proton gets deviated from its original direction is (Proton mass $= 1.6 \times 10^{-27} \ kg$ and charge of proton $= 1.6 \times 10^{-19} \ C$) (in $rad$)