Question

An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience

• A. a force along the axis
• B. a force perpendicular to the axis
• C. no force
• D. a force at an angle of $45^\circ$ to the axis

Hint:

A magnetic field can never exert a force on any moving charge if the charge moves parallel or anti-parallel to the field lines ($\vec{v} \parallel \vec{B}$). Since a circular loop's field points strictly along its axis, any charge moving straight down the center passes through completely unaffected.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An electron moving with a velocity vector is shot directly down the central longitudinal axis of a current-carrying circular ring loop. We need to determine the direction or magnitude of the magnetic force exerted on it.

Step 2: Key Formula or Approach:
1. The magnetic field vector $\vec{B}$ produced at any coordinate point along the central axis of a circular current loop points strictly along that central axis line.
2. The magnetic Lorentz force $\vec{F}$ acting on a moving point charge $q$ inside a magnetic field is defined by the vector cross product:
$$\vec{F} = q(\vec{v} \times \vec{B})$$ The magnitude of this vector cross product simplifies to:
$$F = |q|vB\sin\theta$$ where $\theta$ is the angle separating the velocity vector $\vec{v}$ from the magnetic field vector $\vec{B}$.

Step 3: Detailed Explanation:
Let's analyze the alignment of vectors in this system:
Since the electron is projected along the axis, its velocity vector $\vec{v}$ points parallel to the axis line.
The magnetic field lines $\vec{B}$ generated by the circular current loop also point parallel to the axis line.
Because both vectors point along the exact same linear axis, they are collinear. This means the angle between the velocity vector and the magnetic field vector is either $\theta = 0^\circ$ (same direction) or $\theta = 180^\circ$ (opposite directions). Substitute these angles into the force magnitude equation:
$$\sin(0^\circ) = 0 \quad \text{and} \quad \sin(180^\circ) = 0$$ $$F = |q|vB(0) = 0\ \text{N}$$ Since the cross product of parallel vectors is zero, the magnetic field cannot exert any mechanical force on the moving electron.

Step 4: Final Answer:
The electron will experience no force, which corresponds to option (C).