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

Step 1: Understanding the Concept:
When an electrically charged particle travels through a region of space occupied by a magnetic field, it experiences a deflecting force known as the magnetic component of the Lorentz force.
Unlike the electric force, which acts in the direction of the electric field lines, the magnetic force is inherently three-dimensional and cross-product-based.
This force is always perpendicular to both the velocity vector of the particle and the magnetic field vector.
Because the force is perpendicular to the motion, it can never change the speed (and thus the kinetic energy) of the particle; it can only change the direction of its velocity.
The direction of this force is dictated by the right-hand rule for positive charges.
If the particle's charge is negative, the resulting force direction is flipped by \(180^{\circ}\) (the opposite of the cross product result).
Step 2: Key Formula or Approach:
The mathematical vector equation that defines this physical interaction is given by:
\[ \vec{F}_m = q (\vec{v} \times \vec{B}) \]
To solve for the direction in a coordinate system, we must use the unit vector cross-product properties of a standard right-handed Cartesian system:
- \(\hat{i} \times \hat{j} = \hat{k}\)
- \(\hat{j} \times \hat{k} = \hat{i}\)
- \(\hat{k} \times \hat{i} = \hat{j}\)
And the anti-commutative property: \(\vec{a} \times \vec{b} = - (\vec{b} \times \vec{a})\).
Step 3: Detailed Explanation:
Let's analyze the specific vector components provided in the problem description:
- The charge of the particle is specified as \(q = +q\) (positive).
- The velocity vector is \(\vec{v} = v\hat{i}\) (the particle is moving along the positive X-axis).
- The magnetic field vector is \(\vec{B} = B\hat{j}\) (the field is pointing along the positive Y-axis).
Substituting these components into the Lorentz force equation gives:
\[ \vec{F}_m = (+q) [ (v\hat{i}) \times (B\hat{j}) ] \]
We can pull the scalar magnitudes \(q, v,\) and \(B\) out of the cross product:
\[ \vec{F}_m = qvB (\hat{i} \times \hat{j}) \]
Using the fundamental property of unit vectors where \(\hat{i} \times \hat{j} = \hat{k}\), our expression simplifies to:
\[ \vec{F}_m = qvB \hat{k} \]
This confirms that the force vector points in the direction of the unit vector \(+\hat{k}\), which corresponds to the positive Z-axis.
If the charge had been negative, the coefficient would be \(-q\), making the force point in the \(-\hat{k}\) direction.
Since the charge is positive, the force aligns perfectly with the result of the right-hand rule (\(\hat{i} \times \hat{j}\)).
Physically, as the particle enters the field, it will begin to curve upward into the Z-plane.
Step 4: Final Answer:
The direction of the magnetic Lorentz force acting on the positive charge is \(+\hat{k}\).