Animated Solution for Physics - Magnetic Effects of Current: An electron is moving along +x-direction with a velocity of $6\times 10^6{\text{ ms}}^{-1}$. It enters a region of uniform electric field of $300\text{ V/cm}$ pointing along +y-direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x-direction will be

The problem of a charged particle moving through crossed electric and magnetic fields is a classic setup in physics, famously used by J.J. Thomson to discover the electron! In this problem, we are tasked with finding the exact magnetic field required to keep an electron moving in a perfectly straight line despite the presence of an electric field.

This setup is known as a velocity selector, and it relies on the delicate balance of the Lorentz force. Let's break down the physics step-by-step.

A Dance of Fields

Imagine an electron cruising along the positive x-axis with a staggering velocity of $v=6\times 10^6\text{ m/s}$. Suddenly, it enters a region where a uniform electric field is pointing along the positive y-axis.

The electric field is given as $E=300\text{ V/cm}$. Before we do any physics, we must ensure our units are consistent. Standard SI units require meters, not centimeters. Since there are 100 centimeters in a meter, a change of 300 volts over one centimeter is equivalent to a change of $30,000$ volts over a full meter.

So, our electric field vector is $\mathbf{E}=3\times 10^4\hat{\mathbf{j}}\text{ V/m}$.

As the electron enters this electric field, it immediately feels a force. The electric force on a charged particle is given by the simple equation:

Because the electron carries a negative charge ($-e$), the force it experiences is in the exact opposite direction of the electric field. Since the electric field points up (positive y-direction), the electric force pushes the electron down (negative y-direction).

If this were the only force acting on the electron, it would trace out a parabolic path downwards, much like a ball thrown horizontally under gravity. But the problem states that the electron keeps moving along the x-direction. This implies that the net force on the electron must be zero.

To keep the electron on its straight path, we need a counter-force to perfectly cancel the downward electric force. This is where the magnetic field comes in. We need the magnetic force (${\mathbf{F}}_m$) to point straight up, along the positive y-axis.

The magnetic force on a moving charge is given by the cross product:

We know the velocity is along the positive x-axis ($\mathbf{v}=v\hat{\mathbf{i}}$), and we need the resulting magnetic force to be along the positive y-axis ($+eE\hat{\mathbf{j}}$). Let's set up the equation:

Dividing both sides by $-e$, we get:

Now we must play detective to find the direction of $\mathbf{B}$. We need a vector $\mathbf{B}$ such that when we cross $\hat{\mathbf{i}}$ with it, we get $-\hat{\mathbf{j}}$.

Let's recall our standard unit vector cross products: - $\hat{\mathbf{i}}\times \hat{\mathbf{j}}=\hat{\mathbf{k}}$ - $\hat{\mathbf{i}}\times \hat{\mathbf{k}}=-\hat{\mathbf{j}}$

Bingo! The magnetic field must point along the $\hat{\mathbf{k}}$ direction, which is the positive z-axis.

Let's verify this physically using the right-hand rule. Point your fingers in the direction of velocity (+x). Curl them towards the magnetic field (+z). Your thumb points in the -y direction. This is the direction of $\mathbf{v}\times \mathbf{B}$. However, because the electron is negatively charged, we flip the direction of the force 180 degrees. The actual magnetic force is in the +y direction. Perfect! It exactly opposes the downward electric force.

Now that we have the direction (+z), we just need the magnitude. By equating the magnitudes of the two forces, we get:

Notice how the charge $e$ beautifully cancels out. This means a velocity selector works for any charged particle, regardless of its charge or mass, as long as it has the right velocity!

Solving for $B$:

Now, we substitute our carefully converted values:

To match the options, we adjust the scientific notation:

Combining the magnitude and direction, our final magnetic field vector is:

This corresponds to a magnitude of $5\times 10^{-3}\text{ T}$ along the +z-direction, making option (c) the correct answer. The elegance of this problem lies in how the vector cross product naturally dictates the geometry of the fields required to tame the electron's path!