Question

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

• A. 1.28\( \times \)10\(^{-18}\) N
• B. 12.8\( \times \)10\(^{-18}\) N
• C. 12.8\( \times \)10\(^{-17}\) N
• D. 1.28\( \times \)10\(^{-21}\) N

Hint:

In EM wave problems, remember the simple relationship \(E = cB\). The magnetic force is often much smaller than the electric force on a non-relativistic particle. To find the maximum force, use the amplitude values of the fields and ensure the velocity is perpendicular to the field.

Answer 1

The Correct Option is B

The given problem revolves around electromagnetism, specifically the force exerted on an electron by a magnetic field associated with an electromagnetic wave. To determine the maximum magnetic force on the electron, we can use the following approach:

1. First, understand the context: The electric field of the light beam is given by \(E = 800 \, \text{sin}\left(\omega\left(t - \frac{x}{c}\right)\right)\). Here, the amplitude of the electric field is 800 V/m.
2. Since light is an electromagnetic wave, the magnetic field \(B\) is related to the electric field \(E\) by the relation \(B = \frac{E}{c}\), where \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \, \text{m/s}\).
3. Calculate the amplitude of the magnetic field: 

\(B_0 = \frac{E_0}{c} = \frac{800}{3 \times 10^8} \, \text{Tesla}\)

1. Performing the division, we get: \(B_0 \approx 2.67 \times 10^{-6} \, \text{T}\).
2. The magnetic force \(F\) on a moving charge \(q\) with velocity \(\mathbf{v}\) in a magnetic field \(\mathbf{B}\) is given by the equation: \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\).
3. Since the velocity of the electron is perpendicular to the direction of the magnetic field, the cross product simplifies to: \(F = qvB\).
4. The charge of an electron is \(q = -1.6 \times 10^{-19} \, \text{C}\), and the given speed of the electron is \(v = 3 \times 10^7 \, \text{m/s}\).
5. Substitute the values:

\(F_{\text{max}} = (1.6 \times 10^{-19}) (3 \times 10^7) (2.67 \times 10^{-6}) \, \text{N}\)

1. Simplifying this, we find: \(F_{\text{max}} = 12.8 \times 10^{-18} \, \text{N}\).

Thus, the maximum magnetic force exerted on the electron is 12.8 × 10^-18 N, making the correct answer option 2.