An electron with kinetic energy $$5 \text{ eV}$$ enters a region of uniform magnetic field of $$3 \; \mu T$$ 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 electron moves along the same path, is ______ $$\text{NC}^{-1}$$. (Given, mass of electron $$= 9 \times 10^{-31} \text{ kg}$$, electric charge $$= 1.6 \times 10^{-19} \text{ C}$$)

An electron with KE = 5 eV enters a magnetic field of 3 $$\mu$$T. Find the electric field $$E$$ for the electron to move undeflected.

$$KE = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times KE}{m}}$$

Convert KE to Joules: $$5 \text{ eV} = 5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19}$$ J.

$$v = \sqrt{\frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{1.778 \times 10^{12}}$$

$$v \approx 1.33 \times 10^6 \text{ m/s}$$

$$qE = qvB \implies E = vB$$

$$E = 1.33 \times 10^6 \times 3 \times 10^{-6} = 3.99 \approx 4 \text{ N/C}$$

The correct answer is 4 NC$$^{-1}$$.