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

The provided electric field of an electromagnetic wave in free space is given by:

\[\vec{E} = 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (4\hat{i} - 3\hat{j}) \, \text{N/C}.\]

The objective is to determine the associated magnetic field \( \vec{B} \). The relationship between the electric and magnetic fields in an electromagnetic wave is:

\[\vec{B} = \frac{1}{c} \hat{k} \times \vec{E}\]

where \(c\) is the speed of light (\(3 \times 10^8\) m/s). The wave vector \( \hat{k} \) is derived from the coefficients of \(x\) and \(y\) in the phase term of the electric field, normalized to represent the direction of propagation. Specifically, the phase term is \( - 5 \times 10^{-3} (3x + 4y) \), indicating the wave vector is proportional to \(3\hat{i} + 4\hat{j}\). Normalizing this vector gives the unit wave vector:

\[\vec{k} = \frac{3\hat{i} + 4\hat{j}}{\sqrt{3^2 + 4^2}} = \frac{3\hat{i} + 4\hat{j}}{5}\]

The cross product \( \vec{k} \times \vec{E} \) is calculated as follows:

\[\vec{k} \times \vec{E} = \frac{3\hat{i} + 4\hat{j}}{5} \times 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (4\hat{i} - 3\hat{j})\]

The unit vector part of this product is:

\[\frac{1}{5} (3\hat{i} + 4\hat{j}) \times (4\hat{i} - 3\hat{j}) = \frac{1}{5} \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 4 & -3 & 0 \end{array} \right| = \frac{1}{5} [(0-0)\hat{i} - (0-0)\hat{j} + (3(-3) - 4(4))\hat{k}] = \frac{1}{5}(-9-16)\hat{k} = -5\hat{k}\]

Therefore, the magnetic field is:

\[\vec{B} = \frac{1}{c} \times 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (-5\hat{k})\]

The resulting magnetic field is:

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