91Ó°ÊÓ

The Lorentz force exerted on a charged particle moving in a magnetic field is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \). The particle, with charge \( q \), moves with speed \( v \) in the \(-y\)-direction, so its velocity vector is \( \vec{v} = -v \hat{j} \). The magnetic field vector is \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \). To find the force \( \vec{F} \), compute the cross product \( \vec{v} \times \vec{B} \):\[ \vec{v} \times \vec{B} = (-v \hat{j}) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) \].Calculate each component:- \((-v \hat{j}) \times (B_x \hat{i}) = vB_x \hat{k}\),- \((-v \hat{j}) \times (B_y \hat{j}) = 0\),- \((-v \hat{j}) \times (B_z \hat{k}) = vB_z \hat{i}\).Thus, \( \vec{v} \times \vec{B} = vB_z \hat{i} + vB_x \hat{k} \), and the force is:\[ \vec{F} = q(vB_z \hat{i} + vB_x \hat{k}) = qvB_z \hat{i} + qvB_x \hat{k} \].Therefore, the force components are \( F_x = qvB_z \), \( F_y = 0 \), and \( F_z = qvB_x \).

Given that \( q > 0 \), all components of the force \( \vec{F} \) need to be nonnegative. The force components derived are \( F_x = qvB_z \) and \( F_z = qvB_x \). Since \( q > 0 \) and \( v > 0 \), to ensure \( F_x \geq 0 \) and \( F_z \geq 0 \), both \( B_x \) and \( B_z \) must be positive. Thus, the required signs are:- \( B_x > 0 \)- \( B_z > 0 \)Since \( F_y = 0 \), \( B_y \) does not affect the conditions.

Assuming \( q < 0 \) and \( B_x = B_y = B_z > 0 \), the force components are \( F_x = qvB_z \) and \( F_z = qvB_x \). Since both components depend on the direction of charge, we get:- \( F_x = - |q| v B_x \)- \( F_z = - |q| v B_x \)The force vector \( \vec{F} = - |q| v B_x (\hat{i} + \hat{k}) \). The force direction is along the vector that makes equal angles with \( \hat{i} \) and \( \hat{k} \). This is a diagonal in the \( xz \)-plane, pointing in the negative \( x \text{-} z \) direction. The magnitude is given by:\[ |\vec{F}| = \sqrt{ (-|q| v B_x)^2 + 0^2 + (-|q| v B_x)^2 } = |q|vB_x \sqrt{2} \].