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 experienced by a charged particle in a magnetic field is quantified by the equation: \[ F = qvB \sin \theta \] In this context: - \(q = 1.6 \times 10^{-19} \, \text{C}\) represents the charge of the electron. - \(v = 3 \times 10^6 \, \text{m/s}\) denotes the electron's speed. - \(B = 0.05 \, \text{T}\) is the strength of the magnetic field. - \(\theta = 30^\circ\) signifies the angle between the electron's velocity vector and the magnetic field vector. Upon substitution of these values: \[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (0.05) \times \sin 30^\circ \] Given that \(\sin 30^\circ = 0.5\), the calculation proceeds as: \[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (0.05) \times 0.5 \] The resultant force magnitude is: \[ F = 1.2 \times 10^{-14} \, \text{N} \] Therefore, the magnetic force acting on the electron has a magnitude of \(1.2 \times 10^{-14}\) N.