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 $0.3\, tesla$ at an angle $30^o$ with direction of field. The radius of curvature will be ......$cm$

An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of $3 \mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $\mathrm{E}$, so that electron moves along the same path, is . . . . . $\mathrm{NC}^{-1}$.

(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )

A particle of charge $q$ and mass $m$ is moving along the $x$ -axis with a velocity $v$ and enters a region of electric field $E$ and magnetic field $B$ as shown in figure below for which figure the net force on the charge may be zero

Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

In an experiment, electrons are accelerated, from rest, by applying, a voltage of $500 \,V.$ Calculate the radius of the path if a magnetic field $100\,mT$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19}\,C$ Mass of the electron $= 9.1 \times 10^{-31}\,kg$ ]

A electron $(q = 1.6 \times 10^{-19}\, C)$ is moving at right angle to the uniform magnetic field $3.534 \times 10^{-5}\, T$. The time taken by the electron to complete a circular orbit is......$µs$