Question

The electric field of an electromagnetic wave in free space is represented as \( \vec{E} = E_0 \cos(\omega t - kz) \hat{i} \). The corresponding magnetic induction vector will be:

• A. \( \vec{B} = E_0 C \cos(\omega t - kz) \hat{j} \)
• B. \( \vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j} \)
• C. \( \vec{B} = E_0 C \cos(\omega t + kz) \hat{j} \)
• D. \( \vec{B} = \frac{E_0}{C} \cos(\omega t + kz) \hat{j} \)

Answer 1

The Correct Option is B

The problem asks for the magnetic induction vector \( \vec{B} \) corresponding to a given electric field vector \( \vec{E} \) of an electromagnetic wave in free space.

Concept Used:

For an electromagnetic wave propagating in free space, the following properties hold:

1. The electric field vector \( \vec{E} \), the magnetic field vector \( \vec{B} \), and the direction of wave propagation (\( \vec{k} \)) are mutually perpendicular.
2. The vectors form a right-handed system, meaning the direction of propagation is given by the direction of the cross product \( \vec{E} \times \vec{B} \).
3. The magnitudes of the electric and magnetic fields are related by the speed of light in vacuum, \( C \). The relationship is \( E = C B \), which implies \( B = \frac{E}{C} \).
4. The electric and magnetic fields oscillate in the same phase.

Step-by-Step Solution:

Step 1: Analyze the given electric field vector to determine its properties.

The electric field vector is given by:

\[ \vec{E} = E_0 \cos(\omega t - kz) \hat{i} \]

From this equation, we can deduce:

• The electric field oscillates along the x-axis, as indicated by the unit vector \( \hat{i} \).
• The wave propagates along the positive z-axis. This is determined by the term \( -kz \) inside the cosine function. The direction of propagation is \( \hat{k} \).
• The phase of the wave is \( (\omega t - kz) \).

Step 2: Determine the direction of the magnetic field vector \( \vec{B} \).

The direction of propagation is given by the direction of \( \vec{E} \times \vec{B} \). We know the direction of propagation is \( \hat{k} \) and the direction of \( \vec{E} \) is \( \hat{i} \). Let the direction of \( \vec{B} \) be represented by a unit vector \( \hat{b} \).

\[ \text{Direction}(\vec{E} \times \vec{B}) = \text{Direction of propagation} \] \[ \hat{i} \times \hat{b} = \hat{k} \]

Using the properties of the cross product of Cartesian unit vectors, we know that \( \hat{i} \times \hat{j} = \hat{k} \). Therefore, the direction of the magnetic field vector \( \vec{B} \) must be along the y-axis, so \( \hat{b} = \hat{j} \).

Step 3: Determine the magnitude and phase of the magnetic field vector \( \vec{B} \).

The magnitude of the magnetic field is related to the magnitude of the electric field by \( B = \frac{E}{C} \). The amplitude of the magnetic field, \( B_0 \), will therefore be related to the amplitude of the electric field, \( E_0 \), by:

\[ B_0 = \frac{E_0}{C} \]

The electric and magnetic fields oscillate in the same phase. Since the phase of the electric field is \( (\omega t - kz) \), the phase of the magnetic field must also be \( (\omega t - kz) \).

Final Computation & Result:

Step 4: Assemble the complete magnetic field vector.

Combining the amplitude, phase, and direction found in the previous steps, we can write the expression for the magnetic field vector:

\[ \vec{B} = B_0 \cos(\omega t - kz) \hat{j} \]

Substituting \( B_0 = \frac{E_0}{C} \):

\[ \vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j} \]

Comparing this result with the given options, we find that it matches the second option.

The corresponding magnetic induction vector is \( \vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j} \).

Answer 2

Since \( \vec{B} = \frac{\vec{E}}{C} \times \hat{k} \):

\[ \vec{B} = \frac{E_0}{C} \cos(\omega t - kx) \hat{j} \]