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. \( 2.4 \times 10^{-14} \, \text{N} \)
• B. \( 1.2 \times 10^{-14} \, \text{N} \)
• C. \( 3 \times 10^{-14} \, \text{N} \)
• D. \( 4.8 \times 10^{-14} \, \text{N} \)

Hint:

The magnetic force on a moving charge is given by \( F = qvB \sin \theta \), where \( \theta \) is the angle between the velocity and magnetic field.

Answer 1

The Correct Option is B

The magnetic force \(F\) on a moving charge is quantified by the equation: \[ F = qvB \sin \theta \] With the parameters defined as: - \( q = 1.6 \times 10^{-19} \, \text{C} \), - \( v = 3 \times 10^6 \, \text{m/s} \), - \( B = 0.05 \, \text{T} \), - \( \theta = 30^\circ \). Substituting these values yields: \[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (0.05) \times \sin(30^\circ) \] \[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (0.05) \times 0.5 \] \[ F = 1.2 \times 10^{-14} \, \text{N} \] Consequently, the calculated magnetic force is \( 1.2 \times 10^{-14} \, \text{N} \).