Question

An electron is moving with a velocity \( (2\hat{i} + 3\hat{j}) \) m/s in an electric field \( (3\hat{i} + 6\hat{j} + 2\hat{k}) \) V/m and a magnetic field \( (2\hat{j} + 3\hat{k}) \) T. The magnitude and direction (with x-axis) of the Lorentz force acting on the electron is

• A. \( 9.6 \times 10^{-19} N, \quad \theta = \cos^{-1} \left(\frac{2}{\sqrt{5}}\right) \)
• B. \( 9.6 \times 10^{-19} N, \quad \theta = \cos^{-1} \left(\frac{5}{\sqrt{2}}\right) \)
• C. \( 2.15 \times 10^{-18} N, \quad \theta = \cos^{-1} \left(\frac{2}{\sqrt{5}}\right) \)
• D. \( 2.15 \times 10^{-18} N, \quad \theta = \cos^{-1} \left(\frac{5}{\sqrt{2}}\right) \)

Hint:

Lorentz force includes contributions from both electric and magnetic fields. Use the cross-product formula for motion in a magnetic field.

Answer 1

The Correct Option is C

The Lorentz force is given by: \[ \mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \] Computing the cross product \( \mathbf{v} \times \mathbf{B} \): \[ (2\hat{i} + 3\hat{j}) \times (2\hat{j} + 3\hat{k}) \] Solving for \( \mathbf{F} \), we get: \[ F = 2.15 \times 10^{-18} N \] The direction is given by: \[ \theta = \cos^{-1} \left(\frac{2}{\sqrt{5}}\right) \] Thus, the correct answer is \( 2.15 \times 10^{-18} N, \quad \theta = \cos^{-1} \left(\frac{2}{\sqrt{5}}\right) \).