Step 1: Understanding the Concept:
In a plane electromagnetic wave, the oscillating electric field and magnetic field are always in phase and perpendicular to each other.
The entire wave structure propagates in a direction that is perpendicular to both \(\vec{E}\) and \(\vec{B}\).
The mathematical form of the wave equation is usually \(f(kx - \omega t)\).
Here, \(k = 2\pi/\lambda\) is the wave number and \(\omega = 2\pi\nu\) is the angular frequency.
The ratio of the amplitude of the electric field (\(E_0\)) to the amplitude of the magnetic field (\(B_0\)) is equal to the speed of the wave \(c\).
Key Formula or Approach:
1. Amplitude relation: \(E_0 = c B_0\).
2. Speed relation: \(c = \nu \lambda\).
3. Direction relation: \(\hat{E} \times \hat{B} = \text{Direction of Propagation}\).
Step 2: Detailed Explanation:
1. Finding the Amplitude:
We know \(E_0 = c B_0\).
Substitute \(c = \nu \lambda\) (Frequency \(\times\) Wavelength).
So, \(E_0 = (\nu \lambda) B_0\).
2. Identifying Propagation Direction:
The term in the sine function is \((2\pi\nu t - \frac{2\pi x}{\lambda})\).
This represents a wave moving in the positive \(x\)-direction (since the coefficients of \(t\) and \(x\) have opposite signs).
So, the propagation unit vector is \(+\hat{i}\).
3. Finding Electric Field Direction:
The magnetic field \(\vec{B}\) is given in the \(\hat{j}\) direction.
We need to find \(\hat{E}\) such that \(\hat{E} \times \hat{j} = \hat{i}\).
Using the unit vector cross product rules:
- \(\hat{i} \times \hat{j} = \hat{k}\)
- \(\hat{j} \times \hat{k} = \hat{i}\)
- \(\hat{k} \times \hat{i} = \hat{j}\)
Looking at \(\hat{j} \times \hat{k} = \hat{i}\), this tells us that if \(\vec{B}\) were \(\hat{k}\), \(\vec{E}\) would be \(\hat{j}\).
But here \(\vec{B}\) is \(\hat{j}\). Let's use \(\hat{k} \times \hat{j} = -\hat{i}\).
Multiplying by \(-1\): \((-\hat{k}) \times \hat{j} = + \hat{i}\).
Thus, the electric field direction must be \(-\hat{k}\).
4. Constructing the Final Vector:
The electric field will have the same phase as the magnetic field.
\(\vec{E} = E_0 \sin(\text{phase}) (-\hat{k})\)
\(\vec{E} = -(\nu\lambda B_0) \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}\).
Step 3: Final Answer:
The associated electric field vector is \(-\nu\lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}\).