Question

An electron enters a magnetic field of magnitude 0.05 T at a speed of \(3 \times 10^6\) m/s making an angle of 30° with the field direction. What is the magnitude of magnetic force on it? (Charge of electron = \(1.6 \times 10^{-19}\) C)

• A. \(2.4 \times 10^{-14}\) N
• B. \(1.2 \times 10^{-14}\) N
• C. \(3.0 \times 10^{-14}\) N
• D. \(4.8 \times 10^{-14}\) N

Hint:

The magnetic force is proportional to the charge, velocity, magnetic field strength, and the sine of the angle between the velocity and magnetic field. Always use the right-hand rule to determine the direction of force.

Answer 1

The Correct Option is B

The magnetic force on a charged particle moving in a magnetic field is given by: \[ F = qvB \sin \theta \] Where: - \(q = 1.6 \times 10^{-19} \, \text{C}\) (charge of electron), - \(v = 3 \times 10^6 \, \text{m/s}\) (speed of electron), - \(B = 0.05 \, \text{T}\) (magnetic field), - \(\theta = 30^\circ\) (angle between velocity and magnetic field). Substituting the values: \[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (0.05) \times \sin 30^\circ \] Since \(\sin 30^\circ = 0.5\), we get: \[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (0.05) \times 0.5 \] \[ F = 1.2 \times 10^{-14} \, \text{N} \] Thus, the magnitude of the magnetic force on the electron is \(1.2 \times 10^{-14}\) N.