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

Step 1: Understanding the Question:
We have an electromagnetic wave and an electron moving perpendicular to the wave's propagation direction. We need to find the maximum magnetic force on the electron.
Step 2: Key Formula or Approach:
1. The equation of the electric field gives the amplitude \(E_0\).
2. In an electromagnetic wave, the amplitudes of the electric field (\(E_0\)) and magnetic field (\(B_0\)) are related by \(B_0 = E_0 / c\), where `c` is the speed of light (\(3 \times 10^8\) m/s).
3. The magnetic force on a charge `q` moving with velocity `v` in a magnetic field `B` is given by the Lorentz force formula: \( \vec{F}_m = q(\vec{v} \times \vec{B}) \).
4. The maximum magnetic force occurs when `v` is perpendicular to `B` and the magnetic field is at its maximum value, \(B_0\). The magnitude is \(F_{max} = qvB_0\).
Step 3: Detailed Explanation:
From the given equation for the electric field, \(E = 800 \sin\omega(t - x/c)\), we can identify the amplitude of the electric field as \(E_0 = 800\) V/m.
The wave propagates in the +x direction. The electric field oscillates in a direction perpendicular to it (let's say the y-direction), and the magnetic field oscillates perpendicular to both (in the z-direction).
Calculate the amplitude of the magnetic field, \(B_0\):
\[ B_0 = \frac{E_0}{c} = \frac{800}{3 \times 10^8} \text{ T} \] The electron moves normal to the propagation direction (x-axis) with speed \(v = 3 \times 10^7\) m/s. For the magnetic force to be maximum, the electron's velocity must be perpendicular to the magnetic field. Since `B` is in the z-direction, the velocity can be in the y-direction. The condition is satisfied.
Now, calculate the maximum magnetic force \(F_{max}\). The charge of an electron is \(e = 1.6 \times 10^{-19}\) C.
\[ F_{max} = e v B_0 \] \[ F_{max} = (1.6 \times 10^{-19} \text{ C}) \times (3 \times 10^7 \text{ m/s}) \times \left(\frac{800}{3 \times 10^8 \text{ m/s}}\right) \] The terms \(3 \times 10^7\) and \(3 \times 10^8\) simplify:
\[ F_{max} = (1.6 \times 10^{-19}) \times \left(\frac{1}{10}\right) \times 800 \] \[ F_{max} = 1.6 \times 10^{-19} \times 80 \] \[ F_{max} = 128 \times 10^{-19} \text{ N} \] To match the format of the options, we can write this as:
\[ F_{max} = 12.8 \times 10^{-18} \text{ N} \] Step 4: Final Answer:
The maximum magnetic force exerted on the electron is \(12.8 \times 10^{-18}\) N. This corresponds to option (B).