In a chamber, a uniform magnetic field of 4.5 G(1G = $$10^{-4}$$​T) is maintained. An electron is shot into the field with a speed of 4.8 × $$10^6$$​ m/s normal to the field. Find the radius of the circular orbit? Also obtain the frequency of revolution of the electron in its circular orbit? (e = 1.6 × $$10^{-19}$$​C, $$\text{m}_\text{e}$$​ = 9.1 × $$10^{-31}$$​kg) choose an approximate value

Correct option is C

​$$\textbf{Given:} \\\bullet \, \text{Magnetic field strength: } B = 4.5 \, \text{G} = 4.5 \times 10^{-4} \, \text{T} \\\bullet \, \text{Charge of electron: } e = 1.6 \times 10^{-19} \, \text{C} \\\bullet \, \text{Mass of electron: } m_e = 9.1 \times 10^{-31} \, \text{kg} \\\bullet \, \text{Speed of the electron: } v = 4.8 \times 10^6 \, \text{m/s} \\\textbf{Concept:} \\\text{The electron moves in a circular orbit due to the magnetic force that acts as the centripetal force. We can use the following formula to balance the forces:} \\q v B = \frac{m v^2}{r} \\\text{Where:} \\\bullet \, q \text{ is the charge of the electron,} \\\bullet \, v \text{ is the velocity,} \\\bullet \, B \text{ is the magnetic field strength,} \\\bullet \, m \text{ is the mass of the electron,} \\\bullet \, r \text{ is the radius of the circular orbit.} \\\text{Rearranging the equation to solve for the radius:} \\r = \frac{m v}{q B}$$

$$\textbf{Calculation:} \\\text{Substitute the given values into the equation for } r: \\r = \frac{(9.1 \times 10^{-31}) \times (4.8 \times 10^6)}{(1.6 \times 10^{-19}) \times (4.5 \times 10^{-4})} \\r = 0.06 \, \text{m} = 6 \, \text{cm} \\\text{Thus, the radius of the circular orbit is } r = 6 \, \text{cm}. \\\text{The frequency of revolution } f \text{ is related to the angular velocity } \omega \text{ by the formula:} \\\omega = \frac{v}{r} \\\omega = \frac{(4.8 \times 10^6)}{0.06} = 8 \times 10^7 \, \text{rad/s} \\\text{Now, the frequency } f \text{ is given by:} \\f = \frac{\omega}{2 \pi} = \frac{8 \times 10^7}{2 \pi} = 12.6 \times 10^6 \, \text{Hz} = 12.6 \, \text{MHz}$$​​​