Question

A charged particle is moving in a uniform magnetic field \( \mathbf{B} = 2\hat{i} + 3\hat{j} \) T. If it has an acceleration of \( \mathbf{a} = \alpha\hat{i} - 4\hat{j} \) m/s², then the value of \( \alpha \) will be:

• A. \( 3 \)
• B. \( 6 \)
• C. \( 12 \)
• D. \( 2 \)

Hint:

For charged particles in a magnetic field: 
- Use \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \). 
- Solve using determinant expansion for cross product. - Acceleration components give constraints on velocity components.

Answer 1

The Correct Option is B

Step 1: Understanding the force on a charged particle in a magnetic field. The Lorentz force is given by: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] Since force is also related to acceleration: \[ m\mathbf{a} = q (\mathbf{v} \times \mathbf{B}) \] Thus, \[ \mathbf{a} = \frac{q}{m} (\mathbf{v} \times \mathbf{B}) \] Step 2: Computing the cross product \( \mathbf{v} \times \mathbf{B} \). Let \( \mathbf{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \) and \( \mathbf{B} = 2\hat{i} + 3\hat{j} \), then: \[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ 2 & 3 & 0 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{v} \times \mathbf{B} = \hat{i} (0 v_y - 3 v_z) - \hat{j} (0 v_x - 2 v_z) + \hat{k} (v_x 3 - v_y 2) \] \[ = (-3 v_z) \hat{i} + (2 v_z) \hat{j} + (3 v_x - 2 v_y) \hat{k} \] Since \( \mathbf{a} = \frac{q}{m} (\mathbf{v} \times \mathbf{B}) \), equating components: \[ \alpha = -3 v_z \frac{q}{m}, \quad -4 = 2 v_z \frac{q}{m} \] Solving for \( v_z \): \[ v_z = -\frac{4}{2} \frac{m}{q} = -2 \frac{m}{q} \] Substituting into \( \alpha = -3 v_z \frac{q}{m} \): \[ \alpha = -3 \times (-2) = 6 \] Final Answer: \[ \boxed{6} \]