Animated Solution for Physics - Magnetic Effects of Current: For a positively charged particle moving in a $x-y$ plane initially along the $x$-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond $P$. The curved path is shown in the $x-y$ plane and is found to be non-circular.

Imagine a positively charged particle cruising smoothly along the $x$-axis. Suddenly, it crosses a boundary at point $P$ and enters a region filled with invisible electric and magnetic fields. Its path bends downwards, but interestingly, it remains strictly trapped within the $x-y$ plane. Even more fascinating, the curve it traces is not a perfect circle.

What kind of fields could cause this specific behavior? Let's break down the physics step-by-step.

The most crucial clue is that the particle never leaves the $x-y$ plane. This tells us something profound about the forces acting on it. If there were any net force in the $z$-direction (either pointing towards you or away from you), the particle would spiral out of the screen.

Therefore, we can confidently state our first mathematical constraint:

To find the exact fields, we rely on the Lorentz force equation, which combines the effects of both electric and magnetic fields:

Since the particle is initially moving along the $x$-axis, its velocity vector is simply $\mathbf{v}=v\hat{\mathbf{i}}$. We can now test the given options to see which one satisfies our $F_z=0$ constraint.

Let's test Option (a): $\mathbf{E}=0$ and $\mathbf{B}=b\hat{\mathbf{j}}+c\hat{\mathbf{k}}$.

Notice the $qvb\hat{\mathbf{k}}$ term? That's a force in the $z$-direction! This would push the particle out of the $x-y$ plane. So, Option (a) is incorrect.

By applying the same logic, Option (c) and Option (d) also produce a $z$-component of force because their magnetic fields contain a $\hat{\mathbf{j}}$ component, which, when crossed with the velocity $\hat{\mathbf{i}}$, generates a $\hat{\mathbf{k}}$ force.

Now, let's look at Option (b): $\mathbf{E}=a\hat{\mathbf{i}}$ and $\mathbf{B}=c\hat{\mathbf{k}}+a\hat{\mathbf{i}}$.

Since $\hat{\mathbf{i}}\times \hat{\mathbf{i}}=0$ and $\hat{\mathbf{i}}\times \hat{\mathbf{k}}=-\hat{\mathbf{j}}$, the equation simplifies beautifully to:

There is no $\hat{\mathbf{k}}$ component! The force lies entirely in the $x-y$ plane, perfectly matching our visual observation.

But we have one final piece of the puzzle: why is the path non-circular?

A purely magnetic field perpendicular to the velocity provides a constant centripetal force, resulting in a perfect circle. However, in Option (b), we have an electric force $qa\hat{\mathbf{i}}$ acting parallel to the particle's motion.

This tangential force continuously accelerates the particle, causing its speed $v$ to change. Because the magnetic force in the $y$-direction ($-qvc\hat{\mathbf{j}}$) depends directly on this speed $v$, the magnetic force also changes continuously. A changing speed and a changing perpendicular force mean the radius of curvature is constantly shifting, resulting in a non-circular path.

Everything aligns perfectly with Option (b)!