Question

In free space the electromagnetic-wave equation is given as B = 0.2 cos(wt - kx)$\hat{k}$ T. The total average power passing through a square plate of side 10 cm on the plane y + z = 1 is $\frac{N}{100 \pi^2}$ watt. Find the value of N.

Answer 1

1.5π × 10^13

Answer 2

1. Identify Wave Properties: The given magnetic field $\vec{B} = 0.2 \cos(\omega t - kx)\hat{k}$ T indicates a wave propagating in the $+x$ direction with a peak magnetic field amplitude $B_0 = 0.2$ T. The electric field amplitude is $E_0 = cB_0 = (3 \times 10^8 \text{ m/s}) \times (0.2 \text{ T}) = 6 \times 10^7 \text{ V/m}$. The Poynting vector, representing the direction and magnitude of energy flow, is along the $+x$ direction.

2. Calculate Average Intensity: The average intensity $I_{avg}$ of the electromagnetic wave is given by $I_{avg} = \frac{c B_0^2}{2\mu_0}$. $I_{avg} = \frac{(3 \times 10^8 \text{ m/s}) \times (0.2 \text{ T})^2}{2 \times (4\pi \times 10^{-7} \text{ T m/A})} = \frac{3 \times 10^8 \times 0.04}{8\pi \times 10^{-7}} = \frac{0.12 \times 10^{15}}{8\pi} = \frac{1.5 \times 10^{13}}{\pi} \text{ W/m}^2$.

3. Calculate Average Power through Plate: The square plate has a side of $10 \text{ cm} = 0.1 \text{ m}$, so its area is $A = (0.1 \text{ m})^2 = 0.01 \text{ m}^2$. The problem states the plate is "on the plane $y + z = 1$". However, for power to pass through, the plate must be oriented such that its normal has a component along the direction of wave propagation. Given that the wave propagates along the x-axis, it is implied that the plate is oriented perpendicular to the x-axis (i.e., its normal is along $\hat{i}$) for power to pass through it. The average power $P_{avg}$ passing through the plate is $P_{avg} = I_{avg} \times A$. $P_{avg} = \left( \frac{1.5 \times 10^{13}}{\pi} \text{ W/m}^2 \right) \times (0.01 \text{ m}^2) = \frac{1.5 \times 10^{11}}{\pi} \text{ W}$.

4. Determine N: We are given that the total average power is $\frac{N}{100 \pi^2}$ watt. Equating the two expressions for $P_{avg}$: $\frac{N}{100 \pi^2} = \frac{1.5 \times 10^{11}}{\pi}$ Solving for N: $N = \frac{1.5 \times 10^{11}}{\pi} \times 100 \pi^2 = 1.5 \times 10^{11} \times 100 \pi = 150 \pi \times 10^{11} = 1.5 \pi \times 10^{13}$.

The final answer is 1.5π × 10^13.