An electron with kinetic energy $5\ \text{eV}$ enters a region of uniform magnetic field of $3\ \mu\text{T}$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$, so that the electron moves along the same path, is ______ $\text{N C}^{-1}$. (Given, mass of electron $= 9\times10^{-31}\ \text{kg}$, electric charge $= 1.6\times10^{-19}\ \text{C}$)
Numerical value type. Enter your answer.
Answer: 4
$$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2\times5\times1.6\times10^{-19}}{9\times10^{-31}}} = \frac43\times10^6\ \text{m s}^{-1}$$
For no deflection $eE = evB$:
$$E = vB = \frac43\times10^6\times3\times10^{-6} = 4\ \text{N C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics