Animated Solution for Physics - Magnetic Effects of Current: A magnetic field $\mathbf{B}=B_0\hat{\mathbf{j}}$ exists in the region $a<x<2a$ and $\mathbf{B}=-B_0\hat{\mathbf{j}}$, in the region $2a<x<3a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0\hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like

Imagine you are observing a charged particle as it embarks on a journey through a carefully orchestrated magnetic obstacle course. We have a coordinate system where the $x$ and $z$ axes lie flat on our screen, and the $y$-axis points directly out towards us.

The space is divided into two distinct magnetic regions. In the first region, from $x=a$ to $x=2a$, the magnetic field is $\mathbf{B}=B_0\hat{\mathbf{j}}$, pointing straight out of the screen. In the second region, from $x=2a$ to $x=3a$, the field abruptly reverses its direction to $\mathbf{B}=-B_0\hat{\mathbf{j}}$, pointing directly into the screen. A positive point charge enters this arena at $x=a$, moving purely along the $x$-axis with an initial velocity $\mathbf{v}=v_0\hat{\mathbf{i}}$.

To understand the intricate dance of this particle, we must rely on the Lorentz force law. The magnetic force experienced by a moving charge is given by the cross product:

Because this force is always perpendicular to the particle's velocity, it does absolutely zero work. It cannot speed the particle up or slow it down; it can only change its direction. This constant perpendicular force acts as a centripetal force, forcing the particle to travel along a circular path.

Curving Upwards

Let's freeze time at the exact moment the particle crosses the boundary at $x=a$. Its velocity is $\mathbf{v}=v_0\hat{\mathbf{i}}$, and the magnetic field is $\mathbf{B}=B_0\hat{\mathbf{j}}$. Plugging these into our master equation:

Using the right-hand rule, we know that $\hat{\mathbf{i}}\times \hat{\mathbf{j}}=\hat{\mathbf{k}}$. Therefore, the initial force is ${\mathbf{F}}_1=q{v}_0{B}_0\hat{\mathbf{k}}$. This force points straight up along the positive $z$-axis.

As the particle moves forward, this upward force pulls it off its straight-line path, causing it to curve upwards into the $+z$ region. It traces out a beautiful circular arc until it reaches the boundary at $x=2a$.

Curving Downwards

Crossing the boundary at $x=2a$, the rules of the game suddenly change. The magnetic field flips to $\mathbf{B}=-B_0\hat{\mathbf{j}}$. At this instant, the particle is moving diagonally upwards, meaning its velocity has both a positive $x$-component and a positive $z$-component: $\mathbf{v}=v_x\hat{\mathbf{i}}+v_z\hat{\mathbf{k}}$.

Let's calculate the new force:

Since $\hat{\mathbf{i}}\times \hat{\mathbf{j}}=\hat{\mathbf{k}}$ and $\hat{\mathbf{k}}\times \hat{\mathbf{j}}=-\hat{\mathbf{i}}$, we get:

The crucial detail here is the $-z$ component. The force is now pulling the particle downwards and to the right. This sudden reversal causes the particle to stop curving upwards and begin curving downwards.

Because the two magnetic regions have the exact same width ($a$) and the exact same magnetic field strength ($B_0$), the physics is perfectly symmetric. The particle turned by a specific angle $\theta$ in the first region. In the second region, the reversed field forces it to turn by exactly $-\theta$.

By the time the particle reaches the exit boundary at $x=3a$, it has completely undone its initial rotation, and its velocity is perfectly horizontal once again.

When we trace this entire journey, we see a smooth, S-shaped curve: curving up, passing through an inflection point at $x=2a$, and curving back down to exit horizontally. This elegant trajectory perfectly matches the visual representation in graph (a).