A light beam is described by E = 800 sin(omega t - x/c). An electron is allowed to move normal to th

Q: Why did we use $E_0$ instead of the full equation for $E$?

The question asks for the *maximum* magnetic force. The force is maximum when the magnetic field is at its peak, $B_0$. Since $B_0$ is directly proportional to $E_0$, we only need the peak amplitude from the equation.

Q: Why is the angle $\theta$ in the magnetic force formula $90^\circ$?

The electron moves 'normal to the propagation' of the light beam. In an EM wave, the magnetic field is also perpendicular to the direction of propagation. Thus, the velocity vector and the magnetic field vector are perpendicular to each other, making the angle $90^\circ$.

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

Yes, absolutely! The electric field exerts a force $F_e = qE$. However, the question specifically asks only for the *maximum magnetic force*, so we focus entirely on the magnetic component.

Q: Why are there two correct options?

The values $12.8 \times 10^{-18} \text{ N}$ and $1.28 \times 10^{-17} \text{ N}$ are mathematically identical. They are just written with different decimal placements. In JEE, multiple-choice questions can sometimes have multiple correct formats of the same answer!

Q: What if the electron was moving parallel to the light beam?

If the electron moved parallel to the propagation (x-axis), its velocity would still be perpendicular to the magnetic field (which is on the z-axis). The magnetic force would still be $qvB_0$, but its direction would change!

Question

A light beam is described by $E = 800 \sin\left(\omega t - \frac{x}{c}ight)$. An electron is allowed to move normal to the propagation of light beam with a speed $3 	imes 10^7 	ext{ ms}^{-1}$. What is the maximum magnetic force exerted on the electron?

Options:

(A) $1.28 	imes 10^{-17} 	ext{ N}$

(B) $1.28 	imes 10^{-18} 	ext{ N}$

(C) $12.8 	imes 10^{-17} 	ext{ N}$

(D) $12.8 	imes 10^{-18} 	ext{ N}$

Sigma Percentile · Accepted Answer

### The Setup: A Surfing Electron

Imagine a light beam, which is essentially an electromagnetic wave, traveling through the vastness of space. The equation given to us describes its electric field: 
$$E = 800 \sin\left(\omega t - \frac{x}{c}ight)$$

If you look closely at the $\frac{x}{c}$ term, it reveals a crucial piece of information: the wave is propagating along the positive x-axis. Now, picture an electron shooting past, moving perfectly perpendicular to this light beam. Our mission is to find the maximum magnetic force exerted on this tiny, charged surfer.

### Decoding the Wave Equation

To understand the sheer strength of this wave, we need to compare our given equation with the standard wave equation:
$$E = E_0 \sin(\omega t - kx)$$

The number sitting right in front of the sine function is the amplitude, or the peak electric field. By comparing the two equations, we can clearly see that the peak electric field, $E_0$, is exactly $800 	ext{ V/m}$. This represents the maximum push the electric part of the wave can deliver.

### The Golden Rule of EM Waves

But wait, the question asks for the **magnetic force**! This means we need to find the magnetic field. Do you remember the golden rule of electromagnetic waves? The peak electric field and the peak magnetic field are intimately related by the speed of light, $c$:
$$E_0 = c B_0$$

Let's substitute our known values into this relationship. We plug in $800$ for $E_0$ and $3 	imes 10^8 	ext{ m/s}$ for the speed of light. 
$$B_0 = \frac{800}{3 	imes 10^8} 	ext{ T}$$

We will leave it as a fraction for now. In physics, it's often a smart move to delay division until the very end to make the final calculation cleaner and avoid rounding errors.

### The Magnetic Force Encounter

Now, let's shift our focus back to the electron. The magnetic force on a moving charge is given by the Lorentz force formula:
$$F = q v B \sin(	heta)$$

Since the electron moves normal to the wave's propagation, and the magnetic field is also perpendicular to the propagation, the angle $	heta$ between the velocity and the magnetic field is $90^\circ$. This makes $\sin(90^\circ)$ equal to $1$, giving us the maximum possible force:
$$F_{\max} = e v B_0$$

### The Final Calculation

It's time to bring everything together. Let's substitute the charge of an electron ($1.6 	imes 10^{-19} 	ext{ C}$), its given velocity ($3 	imes 10^7 	ext{ m/s}$), and the peak magnetic field we just found:
$$F_{\max} = (1.6 	imes 10^{-19}) 	imes (3 	imes 10^7) 	imes \left(\frac{800}{3 	imes 10^8}ight)$$

Look closely at the numbers. Do you see how the $3$s in the numerator and denominator are perfectly set up to cancel out? Let's cross them out!
$$F_{\max} = 1.6 	imes 10^{-19} 	imes 10^7 	imes \frac{800}{10^8}$$

Grouping the powers of $10$ together, we get $-19 + 7 - 8 = -20$ in the exponent. Multiplying the remaining numbers, $1.6 	imes 800$, gives us $1280$.
$$F_{\max} = 1280 	imes 10^{-20} 	ext{ N}$$

Finally, adjusting the decimal point to match our options, we can write this as:
$$F_{\max} = 12.8 	imes 10^{-18} 	ext{ N}$$
or equivalently,
$$F_{\max} = 1.28 	imes 10^{-17} 	ext{ N}$$

Looking at our choices, we see that both option (a) and option (d) match our results perfectly. This is a multiple-correct question, so both are the right answers!

Animated Solution for Physics - Electromagnetic Waves: A light beam is described by $E = 800 sin (ω t - \frac{x}{c})$ . An electron is allowed to move normal to the propagation of light beam with a speed $3 \times 1 0^{7} ms^{- 1}$ . What is the maximum magnetic force exerted on the electron?

Select Answer:

Visualized Solution

Visualizing the Setup

Standard Wave Equation

Extracting Peak Electric Field

Relating Electric and Magnetic Fields

Calculating Peak Magnetic Field

Magnetic Force Formula

Substituting Values into Force Equation

Simplifying the Expression

Grouping Powers of 10

Final Multiplication

Adjusting the Decimal

Matching the Options

The Sigma Insight: Characteristics of Electromagnetic Waves

The Setup

Decoding the Wave Equation

The Golden Rule of EM Waves

The Magnetic Force Encounter

The Final Calculation

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