Question

An electron is accelerated through a potential difference of 2 kV and then enters a uniform magnetic field of 0.02 T, in a direction perpendicular to it. Find the radius of the path of the electron in the magnetic field. (Mass of electron = 9.1 x 10^(-31) kg).

Options:
Option 1: 7.13 x 10^(-4) m
Option 2: 5.11 x 10^(-5) m
Option 3: 5.1 x 10^(-4) m
Option 4: 7.5 x 10^(-5) m

Answer 1

The radius of the electron's path in the magnetic field is approximately \(5.11 \times 10^{-5}\) meters, calculated using the given formula and values.

Step-by-step explanation:

To calculate the radius of the path of the electron in the magnetic field, we can use the formula: r = (m * v) / (q * B). Where: r is the radius of the path of the electron, m is the mass of the electron, v is the velocity of the electron, q is the charge of the electron, and B is the magnetic field strength. Plugging in the given values: r = (9.1 x 10^(-31) kg * 10 × 10^4 m/s) / (-1.6 × 10^(-19) C * 0.02 T). Solving this equation will give us r = 5.11 x 10^(-5) m.