Question

An electromagnetic wave travelling in \(x\)-direction is described by the field equation} \[ E_y = 300 \sin \omega\left(t - \frac{x}{c}\right). \] If the electron is restricted to move in \(y\)-direction only with speed \(1.5 \times 10^6\) m/s, then the ratio of maximum electric and magnetic forces acting on the electron is:

• A. \(200\)
• B. \(150\)
• C. \(400\)
• D. \(300\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An electron in an electromagnetic wave experiences an electric force (\(F_e = qE\)) and a magnetic force (\(F_m = q(\vec{v} \times \vec{B})\)). In an electromagnetic wave, the magnitudes of the electric and magnetic field amplitudes are related by the speed of light \(c\).
Step 2: Key Formula or Approach:
1. Electric force: \(F_e = e E\).
2. Magnetic force: \(F_m = e v B\).
3. Field relation: \(E = cB \implies B = E/c\).
Step 3: Detailed Explanation:
The maximum electric force is:
\[ F_{e(max)} = e E_{max} \]
The maximum magnetic force is:
\[ F_{m(max)} = e v B_{max} = e v \left( \frac{E_{max}}{c} \right) \]
The ratio of the maximum forces is:
\[ \text{Ratio} = \frac{F_{e(max)}}{F_{m(max)}} = \frac{e E_{max}}{e v E_{max} / c} \]
\[ \text{Ratio} = \frac{c}{v} \]
Substituting the values \(c = 3 \times 10^8 \text{ m/s}\) and \(v = 1.5 \times 10^6 \text{ m/s}\):
\[ \text{Ratio} = \frac{3 \times 10^8}{1.5 \times 10^6} = 2 \times 10^2 = 200 \].
Step 4: Final Answer:
The ratio of maximum electric and magnetic forces is 200.