The magnetic field vector of an electromagnetic wave is given by B = B0 (i + j)/sqrt(2) cos(kz - wt)

Q: Why do we only care about the magnetic field and not the electric field? Doesn't the EM wave have an electric field too?

Excellent observation! An EM wave indeed has an electric field $\mathbf{E}$ perpendicular to $\mathbf{B}$. The total Lorentz force is $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. However, because the phase $\cos(kz - \omega t)$ evaluates to $-1$ at both positions, both the $\mathbf{E}$ and $\mathbf{B}$ vectors are identical for both charges. Since the entire force vector is multiplied by $q$, the ratio of the magnitudes of the total forces will still simply be $q_1/q_2$. The official solution simplifies this by only looking at the magnetic part, but the ratio holds true for the total force as well!

Q: What if the charges had different velocities?

If the velocities were different, say $\mathbf{v}_1$ and $\mathbf{v}_2$, then the cross product terms $\mathbf{v}_1 \times \mathbf{B}$ and $\mathbf{v}_2 \times \mathbf{B}$ would be different. You would then have to calculate the full vector force for each charge and find the ratio of their magnitudes. The shortcut of just taking the ratio of charges wouldn't work.

Q: How did we know $\cos(3\pi)$ is $-1$?

The cosine function has a period of $2\pi$. This means $\cos(\theta + 2\pi) = \cos(\theta)$. Therefore, $\cos(3\pi) = \cos(\pi + 2\pi) = \cos(\pi) = -1$. In general, the cosine of any odd multiple of $\pi$ is $-1$.

Q: What does the $T$ at the end of the magnetic field equation mean?

The $T$ stands for Tesla, which is the standard SI unit for magnetic field strength. It's just indicating the units of the expression.

Q: Why is the direction of the force along the z-axis?

The velocity is along the x-axis ($\hat{\mathbf{i}}$) and the magnetic field has components along the x and y axes ($\hat{\mathbf{i}} + \hat{\mathbf{j}}$). The cross product $\hat{\mathbf{i}} \times (\hat{\mathbf{i}} + \hat{\mathbf{j}}) = (\hat{\mathbf{i}} \times \hat{\mathbf{i}}) + (\hat{\mathbf{i}} \times \hat{\mathbf{j}}) = 0 + \hat{\mathbf{k}}$. So the resulting magnetic force points along the z-axis.

Question

The magnetic field vector of an electromagnetic wave is given by $\mathbf{B} = B_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} \cos(kz - \omega t)	ext{ T}$ where $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ represents unit vector along X and Y-axis respectively. At $t = 0$, two electric charges $q_1$ of $4\pi	ext{ C}$ and $q_2$ of $2\pi	ext{ C}$ located at $\left(0, 0, \frac{\pi}{k}ight)$ and $\left(0, 0, \frac{3\pi}{k}ight)$ respectively, have the same velocity of $0.5 c \hat{\mathbf{i}}$. (where, $c$ is the velocity of light). The ratio of the force acting on charge $q_1$ to $q_2$ is

Options:

(A) $2\sqrt{2} : 1$

(B) $1 : \sqrt{2}$

(C) $2 : 1$

(D) $\sqrt{2} : 1$

Sigma Percentile · Accepted Answer

The problem presents us with an electromagnetic wave and asks for the ratio of forces acting on two moving charges. At first glance, it might seem like we need to dive into a messy cross-product calculation for both charges. But let's take a step back and look at the physics.

### Analyzing the Setup

We are given the magnetic field vector of the wave:
$$ \mathbf{B} = \frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \cos(kz - \omega t) $$

We have two charges, $q_1 = 4\pi	ext{ C}$ and $q_2 = 2\pi	ext{ C}$, located on the z-axis at $z_1 = \frac{\pi}{k}$ and $z_2 = \frac{3\pi}{k}$ respectively. At time $t=0$, both are moving with the exact same velocity $\mathbf{v} = 0.5c\hat{\mathbf{i}}$.

### Decoding the Phase

The key to unlocking this problem without tedious math lies in the phase of the wave, $(kz - \omega t)$. Let's evaluate the magnetic field at the exact locations of our charges at $t=0$.

For the first charge $q_1$ at $z = \frac{\pi}{k}$:
$$ \mathbf{B}_1 = \frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \cos\left(k\frac{\pi}{k} - 0ight) = \frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \cos(\pi) $$
Since $\cos(\pi) = -1$, we get:
$$ \mathbf{B}_1 = -\frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) $$

Now, let's look at the second charge $q_2$ at $z = \frac{3\pi}{k}$:
$$ \mathbf{B}_2 = \frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \cos\left(k\frac{3\pi}{k} - 0ight) = \frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \cos(3\pi) $$
Since $\cos(3\pi)$ is also $-1$, we find:
$$ \mathbf{B}_2 = -\frac{B_0}{\sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) $$

**Notice the magic here:** $\mathbf{B}_1 = \mathbf{B}_2$. Both charges are experiencing the exact same magnetic field vector!

### The Total Lorentz Force

The total force on a moving charge in an electromagnetic field is given by the Lorentz force equation:
$$ \mathbf{F} = q(\mathbf{E} + \mathbf{v} 	imes \mathbf{B}) $$

You might wonder, what about the electric field $\mathbf{E}$? In an electromagnetic wave, the electric field is intrinsically linked to the magnetic field. Because the magnetic fields are identical at both locations, the electric fields must also be identical ($\mathbf{E}_1 = \mathbf{E}_2$).

Since both charges also share the exact same velocity vector $\mathbf{v}$, the entire vector expression inside the parentheses $(\mathbf{E} + \mathbf{v} 	imes \mathbf{B})$ is identical for both $q_1$ and $q_2$.

### The Final Ratio

Because the field and velocity terms are identical, the total force is strictly proportional to the magnitude of the charge:
$$ \mathbf{F} \propto q $$

Therefore, the ratio of the forces is simply the ratio of the charges:
$$ \frac{F_1}{F_2} = \frac{q_1}{q_2} = \frac{4\pi}{2\pi} = \frac{2}{1} $$

The ratio is **$2:1$**. By understanding the spatial periodicity of the wave, we bypassed a complex vector calculation and arrived at the elegant truth of the problem.

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Visualized Solution

$Analyzing the Setup$

$Magnetic Field at q_{1}$

$Evaluating B_{1}$

$Magnetic Field at q_{2}$

$Evaluating B_{2}$

$The Total Lorentz Force$

$Ratio of Forces$

$What if the positions were different?$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

Decoding the Phase

The Total Lorentz Force

The Final Ratio

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field at q1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating B1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field at q2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating B2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Total Lorentz Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the positions were different?