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 problem gives us the electric field of an electromagnetic wave in free space. We need to determine the associated magnetic field. The electric field is given by:

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

We need to find the associated magnetic field \( \vec{B} \). The relationship between the electric field \( \vec{E} \) and the magnetic field \( \vec{B} \) in an electromagnetic wave is given by: 

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

Here, \(c\) is the speed of light in free space, approximately \(3 \times 10^8\) m/s. Since the wave propagates in the direction perpendicular to the electric field direction, we apply the cross product using the factor \( \hat{k} \), the unit vector in the direction of wave propagation.

The given electric field direction is \( (4\hat{i} - 3\hat{j}) \). The wave vector \( \hat{k} \) in the propagation direction can be determined by normalizing the coefficients in the cosine term:

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

Now solve for the \( \vec{k} \times \vec{E} \) which will be normal to \( \vec{E} \) and \( \vec{k} \). Performing the cross product:

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

Simplifying gives:

\[\vec{k} \times \vec{E} = -\left(\frac{9}{5} + \frac{16}{5}\right)\hat{k} = -5\hat{k}\]

So the magnetic field \( \vec{B} \) is:

\[\vec{B} = -\frac{1}{c} \vec{k} \times \vec{E} = -\frac{1}{3 \times 10^8} (57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right]) (5\hat{k})\]

Thus the correct answer is:

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

Answer 2

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:

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

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