Question

A particle carrying a positive charge $q$ moves with a velocity vector $\vec{v} = v\hat{i}$ into a uniform magnetic field region specified by $\vec{B} = B\hat{j}$. What is the vector direction of the magnetic Lorentz force acting on this particle at that instant?}

• A. \( +\hat{k} \)
• B. \( -\hat{k} \)
• C. \( +\hat{i} \)
• D. \( -\hat{j} \)

Hint:

You can verify cross-product directions using a simple clockwise circle memory aid: if you go from \(\hat{i} \rightarrow \hat{j}\), you get positive \(\hat{k}\). If you reverse the order to \(\hat{j} \rightarrow \hat{i}\), you get negative \(-\hat{k}\). Always check the sign of the charge first!

Answer 1

The Correct Option is A

Concept: When an electrically charged particle moves through a magnetic field, it experiences a deflecting force known as the magnetic component of the Lorentz Force. This force relies fundamentally on the relative orientation of the particle's velocity vector and the surrounding magnetic field lines. The exact vector equation mapping this physical interaction is given by: \[ \vec{F} = q(\vec{v} \times \vec{B}) \] This mathematical cross product dictates two critical physical rules:
• The resulting magnetic force vector \(\vec{F}\) is always simultaneously perpendicular to both the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\).
• The direction depends strictly on the sign of the net charge \(q\). If the charge is positive, the force matches the direction of \(\vec{v} \times \vec{B}\). If the charge is negative, the force points in the exact opposite direction (\(-\vec{v} \times \vec{B}\)).

Step 1: Substitute the given vector components into the Lorentz law. Let's look at our given values:
• Net charge of the particle = \(+q\) (positive sign)
• Velocity vector \(\vec{v} = v\hat{i}\) (moving along the positive x-axis)
• Magnetic field vector \(\vec{B} = B\hat{j}\) (pointing along the positive y-axis) Substituting these into the cross-product equation: \[ \vec{F} = q \cdot (v\hat{i} \times B\hat{j}) \]

Step 2: Use scalar and vector properties to evaluate the cross product. Pull the scalar magnitudes (\(v\) and \(B\)) out in front of the vector cross product: \[ \vec{F} = qvB \cdot (\hat{i} \times \hat{j}) \] According to the properties of orthogonal unit vectors in a right-handed Cartesian coordinate system, the cross product of the unit vector along the x-axis (\(\hat{i}\)) and the unit vector along the y-axis (\(\hat{j}\)) yields the positive unit vector along the z-axis (\(\hat{k}\)): \[ \hat{i} \times \hat{j} = \hat{k} \] Substituting this back into our expression gives: \[ \vec{F} = qvB\hat{k} \] Since the unit vector is \(\hat{k}\), the mechanical force points straight along the positive z-direction.