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

To determine the magnetic field \(\vec{B}\) associated with the given electric field \(\vec{E}\) in an electromagnetic wave, we use the relationship between the electric and magnetic fields in a plane wave in free space. The magnitude of the magnetic field \(\vec{B}\) is given by \(|\vec{B}| = \frac{|\vec{E}|}{c}\), where \(c\) is the speed of light, approximately \(3 \times 10^8\) m/s.

Given the electric field:

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

The magnitude of \(\vec{E}\) is 57 N/C.

The corresponding magnitude of \(\vec{B}\) is:

\[|\vec{B}| = \frac{57}{3 \times 10^8} \, \text{T}\]

The direction of \(\vec{B}\) is perpendicular to both \(\vec{E}\) and the wave vector \(\vec{k}\). The wave propagates in the direction of \(\vec{k} = 3\hat{i} + 4\hat{j}\), perpendicular to both \(\vec{E}\) and the direction of propagation. The direction of \(\vec{B}\) can be determined using the right-hand rule. Thus, \(\vec{B}\) is directed along \(\hat{k}\).

However, using the right-hand rule considering the cross product \(\vec{k} \times \vec{E}\), \(\vec{B}\) points in the negative \(\hat{k}\) direction, hence introducing the negative sign:

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

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