An electron (mass $= 9 \times 10^{-31} \, kg$,charge $= 1.6 \times 10^{-19} \, C$) with a kinetic energy of $7.2 \times 10^{-18} \, J$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \, Wb/m^2$. The radius of the orbit is ..... $cm$.

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?

$A$ particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction,extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

Two particles $A$ and $B$ having equal charges $+6\,C$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $2\,cm$ and $3\,cm$ respectively. The ratio of mass of $A$ to that of $B$ is

$A$ charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when