Let's break down the problem step by step: ### **Given:** \[ \mathbf{E} = 10\sqrt{2} \cos(10^8 t + y) \mathbf{a}_x \ \text{V/m} \] We are to find the **instantaneous form of the magnetic field** \(\mathbf{H}\). --- ## **Step 1: Understand the Wave Propagation** The electric field is along \(\mathbf{a}_x\) and the argument of the cosine is \((10^8 t + y)\), so the wave is propagating in the **negative y-direction**. --- ## **Step 2: Relationship Between E and H** For a plane electromagnetic wave in free space (lossless region): \[ \mathbf{H} = \frac{1}{\eta} (\mathbf{a}_k \times \mathbf{E}) \] where: - \(\eta\) = intrinsic impedance of free space (\( \approx 120\pi \ \Omega \)) - \(\mathbf{a}_k\) = unit vector in the direction of propagation (here, \(-\mathbf{a}_y\)) --- ## **Step 3: Calculate the Cross Product** \[ \mathbf{a}_k = -\mathbf{a}_y \] \[ \mathbf{E} = E_x \mathbf{a}_x \] \[ \mathbf{a}_k \times \mathbf{a}_x = (-\mathbf{a}_y) \times \mathbf{a}_x = -(\mathbf{a}_y \times \mathbf{a}_x) \] Recall: \(\mathbf{a}_y \times \mathbf{a}_x = -\mathbf{a}_z\), so \[ -\mathbf{a}_z \] But since it's already negative: \(-(-\mathbf{a}_z) = \mathbf{a}_z\) So the direction of \(\mathbf{H}\) is **\(\mathbf{a}_z\)**. --- ## **Step 4: Find the Magnitude** \[ |\mathbf{E}| = 10\sqrt{2} \cos(10^8 t + y) \] \[ |\mathbf{H}| = \frac{|\mathbf{E}|}{\eta} \] \(\eta = 120\pi\) \[ |\mathbf{H}| = \frac{10\sqrt{2}}{120\pi} \cos(10^8 t + y) \] \[ = \frac{\sqrt{2}}{12\pi} \cos(10^8 t + y) \] --- ## **Step 5: Write the Final Expression** \[ \mathbf{H} = \frac{\sqrt{2}}{12\pi} \cos(10^8 t + y) \mathbf{a}_z \quad \text{A/m} \] --- ## **Step 6: Check Options** **Option (a):** \[ \mathbf{H} = \frac{\sqrt{2}}{12\pi} \cos(10^8 t + y) \mathbf{a}_z \ \text{A/m} \] This matches our result. --- ## **Final Answer:** \[ \boxed{\text{a. } \mathbf{H} = \frac{\sqrt{2}}{12\pi} \cos(10^8 t + y) \mathbf{a}_z \ \text{A/m}} \]