The magnetic field accompanying an electric field in a plane electromagnetic wave is determined by the equation \(E = cB\), where \(E\) represents the electric field magnitude, \(B\) the magnetic field magnitude, and \(c\) the speed of light (approximately \(3 \times 10^8 \text{ m/s}\)).
Given the electric field vector as \( \vec{E} = 9.6 \hat{j} \, \text{V/m} \), and knowing the wave propagates in the X-direction, the magnetic field must be perpendicular to both the direction of propagation (X) and the electric field (Y). Consequently, the magnetic field lies along the Z-direction.
The magnitude of the magnetic field is calculated as:
\(B = \frac{E}{c} = \frac{9.6}{3 \times 10^8}\)
This yields:
\(B = 3.2 \times 10^{-8} \, \text{T}\)
Considering the Z-direction, the magnetic field vector is:
\(\vec{B} = 3.2 \times 10^{-8} \hat{k} \, \text{T}\)
This matches the correct option.
Alternative options are incorrect because:
Therefore, the magnetic field at this point is \(3.2 \times 10^{-8} \hat{k} \, \text{T}\).
A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?
