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:
- \( \vec{B} = E_0 C \cos(\omega t - kz) \hat{j} \)
- \( \vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j} \)
- \( \vec{B} = E_0 C \cos(\omega t + kz) \hat{j} \)
- \( \vec{B} = \frac{E_0}{C} \cos(\omega t + kz) \hat{j} \)
The Correct Option is B
Approach Solution - 1
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:
- The electric field vector \( \vec{E} \), the magnetic field vector \( \vec{B} \), and the direction of wave propagation (\( \vec{k} \)) are mutually perpendicular.
- The vectors form a right-handed system, meaning the direction of propagation is given by the direction of the cross product \( \vec{E} \times \vec{B} \).
- 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 \( B = \frac{E}{C} \).
- 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 \( \vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j} \).
Approach Solution -2
Since \( \vec{B} = \frac{\vec{E}}{C} \times \hat{k} \):
\[ \vec{B} = \frac{E_0}{C} \cos(\omega t - kx) \hat{j} \]
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