Step 1: Understanding the Concept:
Electromagnetic waves are a manifestation of oscillating electric and magnetic fields that propagate through space.
One of the most defining characteristics of these waves is their transverse nature, meaning that both the electric field vector (\(\vec{E}\)) and the magnetic field vector (\(\vec{B}\)) are perpendicular to each other and to the direction of wave propagation (\(\vec{v}\)).
In free space (vacuum), the magnitudes of these fields are intrinsically linked by the speed of light (\(c\)), which is approximately \(3 \times 10^8\) m/s.
The fundamental relationship is expressed as \(E = cB\), highlighting that the electric component is numerically much larger than the magnetic component in SI units.
Furthermore, the direction of these vectors follows a specific right-hand rule convention.
The direction of propagation is given by the cross product \(\vec{E} \times \vec{B}\).
This ensures that the energy flow, represented by the Poynting vector, aligns with the direction of travel.
Key Formula or Approach:
To solve this, we apply two separate criteria:
1. The magnitude relationship: \[ E = cB \]
2. The directional relationship using unit vectors: \[ \hat{E} \times \hat{B} = \hat{c} \]
Where \(\hat{c}\) is the unit vector in the direction of wave propagation.
Step 2: Detailed Explanation:
Calculation of Magnitude:
We are given the magnetic field magnitude \(B = 2 \times 10^{-7}\) T and the wave is in free space.
The speed of light \(c\) is a constant \(3 \times 10^8\) m/s.
Using the magnitude formula:
\[ E = (3 \times 10^8 \text{ m/s}) \times (2 \times 10^{-7} \text{ T}) \]
Multiplying the coefficients: \(3 \times 2 = 6\).
Handling the powers of ten: \(10^8 \times 10^{-7} = 10^1\).
Thus, \(E = 6 \times 10 = 60\) V/m.
Determination of Direction:
The question states the wave travels along the \(x\)-direction, so the propagation unit vector is \(\hat{i}\).
The magnetic field is given as \(\vec{B} = B_y \hat{j}\), so its unit vector is \(\hat{j}\).
Let the electric field direction be represented by a unit vector \(\hat{n}\).
According to the cross product rule: \(\hat{n} \times \hat{j} = \hat{i}\).
From the cyclic properties of unit vectors (\(\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}\)):
We know \(\hat{k} \times \hat{i} = \hat{j}\) and \(\hat{j} \times \hat{k} = \hat{i}\).
However, we need \(\hat{n} \times \hat{j} = \hat{i}\).
Looking at the anti-cyclic property: \(\hat{j} \times \hat{i} = -\hat{k}\) and \(\hat{k} \times \hat{j} = -\hat{i}\).
Multiplying both sides of \(\hat{k} \times \hat{j} = -\hat{i}\) by \(-1\), we get \((-\hat{k}) \times \hat{j} = \hat{i}\).
Therefore, the direction of the electric field must be along the negative \(z\)-axis, represented by \(-\hat{k}\).
Step 3: Final Answer:
Combining the magnitude and direction, the electric field vector is \(\vec{E} = -60\hat{k}\) V/m.