Question

A particle of mass \( m \) and charge \( q \) is moving with velocity \( \mathbf{v} = v_x \hat{i} + v_y \hat{j} \). If it is subjected to a magnetic field \( \mathbf{B} = B_0 \hat{i} \), it will move in a:

• A. straight line path
• B. circular path
• C. helical path
• D. parabolic path

Hint:

A charged particle moves in a helical path when its velocity has a component parallel and perpendicular to the magnetic field.

Answer 1

The Correct Option is C

Motion of a Charged Particle in a Magnetic Field
- The Lorentz Force on a charged particle moving in a magnetic field is: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] - Given \( \mathbf{B} = B_0 \hat{i} \) and \( \mathbf{v} = v_x \hat{i} + v_y \hat{j} \), we calculate the cross-product: \[ \mathbf{F} = q [(v_x \hat{i} + v_y \hat{j}) \times (B_0 \hat{i})] \] - Since \( \hat{i} \times \hat{i} = 0 \) and \( \hat{j} \times \hat{i} = -\hat{k} \), we get: \[ \mathbf{F} = -q v_y B_0 \hat{k} \] - The force acts perpendicular to the velocity component in the \( yz \)-plane, causing circular motion in that plane. - The velocity component along \( x \)-direction remains unchanged, leading to a helical path. Thus, the correct answer is (C) helical path.