An electron is constrained to move along the Y-axis with a speed of 0.1 c (c is the speed of light)

Q: Why is the angle between the velocity and the magnetic field exactly 90 degrees?

In an electromagnetic wave, the electric field, magnetic field, and the direction of propagation are all mutually perpendicular. Since the wave propagates along the x-axis and the electric field is along the y-axis, the magnetic field must be along the z-axis. The electron is moving along the y-axis. The angle between the y-axis and z-axis is 90 degrees.

Q: Why didn't we use the full sine wave equation for the magnetic field?

The question specifically asks for the *maximum* magnetic force. The magnetic force is maximum when the magnetic field is at its peak amplitude ($B_{\text{max}}$). Therefore, we only need the amplitude, not the instantaneous value at a specific time or position.

Q: Does the electric field also exert a force on the electron?

Yes, absolutely! The electric field exerts an electric force $\mathbf{F}_e = q\mathbf{E}$. However, the question explicitly asked only for the maximum *magnetic* force, so we isolated that specific component of the total Lorentz force.

Q: Why did the speed of light 'c' cancel out in the final calculation?

This is a beautiful consequence of the setup. The electron's speed was given as a fraction of the speed of light ($0.1c$), and the magnetic field amplitude is inversely proportional to the speed of light ($E_{\text{max}}/c$). When you multiply them together in the force equation, the 'c' terms naturally cancel out.

Q: What does the negative sign in the phase $(1.5 \times 10^7 t - 5 \times 10^{-2} x)$ mean?

The standard form of a wave equation is $\sin(\omega t - kx)$ or $\sin(kx - \omega t)$. A negative sign between the time and space variables indicates that the wave is traveling in the positive direction of the spatial axis (in this case, the positive x-axis).

Question

An electron is constrained to move along the Y-axis with a speed of $0.1 c$ ($c$ is the speed of light) in the presence of electromagnetic wave, whose electric field is $\mathbf{E} = 30 \hat{\mathbf{j}} \sin(1.5 	imes 10^7 t - 5 	imes 10^{-2} x) 	ext{ V/m}$. The maximum magnetic force experienced by the electron will be (Given, $c = 3 	imes 10^8 	ext{ ms}^{-1}$ and electron charge $= 1.6 	imes 10^{-19} 	ext{ C}$)

Options:

(A) $2.4 	imes 10^{-18} 	ext{ N}$

(B) $4.8 	imes 10^{-19} 	ext{ N}$

(C) $1.6 	imes 10^{-19} 	ext{ N}$

(D) $3.2 	imes 10^{-18} 	ext{ N}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine an electron moving straight up along the $y$-axis. It's moving incredibly fast, at $10\%$ the speed of light, so its velocity is $\mathbf{v} = 0.1c \hat{\mathbf{j}}$.

This electron isn't just moving in empty space; it's bathed in an electromagnetic wave. The problem gives us the electric field equation: $\mathbf{E} = 30 \hat{\mathbf{j}} \sin(1.5 	imes 10^7 t - 5 	imes 10^{-2} x) 	ext{ V/m}$.

From this equation, we can extract two crucial pieces of information. First, the amplitude of the electric field, $E_{	ext{max}}$, is $30 	ext{ V/m}$. Second, the negative sign in the phase $(kx - \omega t)$ tells us the wave is traveling along the positive $x$-axis.

### The Magnetic Field Connection

In any electromagnetic wave, the electric and magnetic fields are tightly linked. They oscillate perpendicular to each other and perpendicular to the direction of wave propagation.

Since the electric field $\mathbf{E}$ is along the $y$-axis and the wave travels along the $x$-axis, the magnetic field $\mathbf{B}$ must oscillate along the $z$-axis.

Furthermore, their amplitudes are related by the speed of light, $c$. The maximum magnetic field is simply the maximum electric field divided by the speed of light: 
$$B_{	ext{max}} = \frac{E_{	ext{max}}}{c}$$

### The Lorentz Force

Our moving electron will feel a magnetic force. The formula for this is given by the Lorentz force law: 
$$\mathbf{F} = q(\mathbf{v} 	imes \mathbf{B})$$

Because the electron's velocity is along the $y$-axis and the magnetic field is along the $z$-axis, the angle $	heta$ between them is exactly $90^\circ$.

This makes our calculation much simpler because $\sin(90^\circ) = 1$. The maximum magnetic force is just the product of the charge, the speed, and the maximum magnetic field:
$$F_{	ext{max}} = q v B_{	ext{max}} \sin 90^\circ = q v B_{	ext{max}}$$

### The Elegant Cancellation

Now, let's substitute $B_{	ext{max}}$ with $\frac{E_{	ext{max}}}{c}$ to get our master equation:
$$F_{	ext{max}} = \frac{q v E_{	ext{max}}}{c}$$

Time to plug in the values! The charge of an electron $q$ is $1.6 	imes 10^{-19} 	ext{ C}$. The velocity $v$ is $0.1 c$. And $E_{	ext{max}}$ is $30 	ext{ V/m}$.

$$F_{	ext{max}} = \frac{(1.6 	imes 10^{-19}) 	imes (0.1 c) 	imes 30}{c}$$

Notice how the speed of light, $c$, is in both the numerator and the denominator. It cancels out beautifully!

$$F_{	ext{max}} = 1.6 	imes 10^{-19} 	imes 0.1 	imes 30$$

Zero point one times thirty is three. And three times one point six is four point eight.

$$F_{	ext{max}} = 4.8 	imes 10^{-19} 	ext{ N}$$

So, the maximum magnetic force experienced by the electron is $4.8 	imes 10^{-19} 	ext{ N}$.

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Visualized Solution

$Visualizing the Setup$

$Analyzing the Electric Field$

$Magnetic Field Amplitude$

$Magnetic Force Formula$

$Maximum Magnetic Force$

$Substituting Values$

$Final Calculation$

The Sigma Insight: Properties of EM Waves

Visualizing the Setup

The Magnetic Field Connection

The Lorentz Force

The Elegant Cancellation

Similar Questions

An electromagnetic wave of frequency $5 GHz$ , is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is ...... $\times 1 0^{7} m/s$ .

An electromagnetic wave of frequency $3 GHz$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be ......... $\times 1 0^{- 2} cm$ .

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Visualized Solution

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

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

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

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

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

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

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