Question

The electric field of an electromagnetic wave in free space is \[ \vec{E} = 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] \left( 4\hat{i} - 3\hat{j} \right) \, \text{N/C}. \] The associated magnetic field in Tesla is:

• A. \( \vec{B} = \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k}) \)
• B. \( \vec{B} = \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (\hat{k}) \)
• C. \( \vec{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k}) \)
• D. \( \vec{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (\hat{k}) \)

Hint:

In problems involving electromagnetic waves, use the relationship between the electric field \( \mathbf{E} \) and magnetic field \( \mathbf{B} \) given by the cross product, and remember that the magnetic field will be perpendicular to both \( \mathbf{E} \) and the wave vector \( \mathbf{K} \).

Answer 1

The Correct Option is C

We are given the electric field of an electromagnetic wave as: \[ \mathbf{E} = 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] \left( 4\hat{i} - 3\hat{j} \right) \, \text{N/C}. \] The relationship between the electric and magnetic fields in an electromagnetic wave is given by the equation: \[ \mathbf{E} \times \mathbf{B} = \mathbf{K}, \] where \( \mathbf{K} \) is the wave vector. The wave vector can be written as: \[ \mathbf{K} = 3\hat{i} + 4\hat{j}. \] Thus, we can calculate \( \mathbf{B} \), the magnetic field, using the cross product: \[ \mathbf{B} = \frac{\mathbf{K} \times \mathbf{E}}{c}, \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light in a vacuum. From the problem, the electric field vector is: \[ \mathbf{E} = \frac{4\hat{i} - 3\hat{j}}{5}. \] The cross product of \( \mathbf{K} \) and \( \mathbf{E} \) gives the direction of the magnetic field. Since \( \mathbf{K} = 3\hat{i} + 4\hat{j} \), the resulting magnetic field vector \( \mathbf{B} \) will be in the \( \hat{k} \)-direction. Thus, the magnetic field is: \[ \mathbf{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k}). \] Thus, the correct answer is option (3).