Question

An electron moving along positive x-direction with velocity \( 2 \times 10^5 \) ms\(^{-1}\) enters a magnetic field of \( \vec{B} = \hat{i} + 4 \hat{j} + 3 \hat{k} \) T. The magnitude of the force acting on the electron is:

• A. \( 1.6 \times 10^{13} \, \text{N} \)
• B. \( 1.6 \times 10^{-13} \, \text{N} \)
• C. \( 1.6 \times 10^{-14} \, \text{N} \)
• D. \( 1.6 \times 10^{-3} \, \text{N} \)

Hint:

The force on a charged particle moving in a magnetic field is calculated using the formula \( F = q (\vec{v} \times \vec{B}) \), where \( q \) is the charge, \( \vec{v} \) is the velocity, and \( \vec{B} \) is the magnetic field.

Answer 1

The Correct Option is B

The force on a charged particle moving in a magnetic field is given by the equation \( F = q (\vec{v} \times \vec{B}) \). Using the given values for velocity \( \vec{v} \) and magnetic field \( \vec{B} \), we calculate the cross product and find the magnitude of the force acting on the electron as \( 1.6 \times 10^{-13} \, \text{N} \).