The magnetic field in a radio wave traveling through air has amplitude $2.5 \times 10^{-11} \mathrm{T}\( and frequency \)3.0 \mathrm{MHz}$. (a) Find the amplitude and frequency of the electric field. (b) The wave is traveling in the \(-y\) -direction. At \(y=0\) and \(t=0\), the magnetic field is $1.5 \times 10^{-11} \mathrm{T}\( in the \)+z$ -direction. What are the magnitude and direction of the electric field at \(y=0\) and \(t=0 ?\)

To find the amplitude and frequency of the electric field, we will use the relationship between the electric and magnetic field amplitudes in an electromagnetic wave: E = B * c where E is the electric field amplitude, B is the magnetic field amplitude (\(2.5 \times 10^{-11} \mathrm{T}\)), and c is the speed of light in a vacuum (\(3.0 \times 10^8 \mathrm{m/s}\)). E = (2.5 × 10^(-11) T) × (3.0 × 10^(8) m/s) = 7.5 × 10^(-3) V/m The frequency of the electric field is the same as the frequency of the magnetic field, which is given as \(3.0 \mathrm{MHz}\). So, the amplitude of the electric field is \(7.5 \times 10^{-3} \mathrm{V/m}\), and the frequency is \(3.0 \mathrm{MHz}\).

Given that the magnetic field is initially \(1.5 \times 10^{-11} \mathrm{T}\) in the \(+z\) direction, we will use Faraday's law to find the magnitude and direction of the electric field at \(y=0\) and \(t=0\). Faraday's law states: \(E = - \frac{d\Phi_B}{dt}\) Where E is the electric field, and \(\Phi_B\) is the magnetic flux. For a radio wave, we have: \(\Phi_B = B \times A\) Where A is the area perpendicular to the magnetic field lines. Since we know the magnetic field has a sinusoidal behavior, we can write: \(B = B_0 \cdot \sin(2 \pi f t)\) Where \(B_0 = 2.5 \times 10^{-11} \mathrm{T}\) is the maximum value of the magnetic field, f is the frequency (3 MHz) and t is the time. Furthermore, at \(y = 0\) and \(t=0\), we have: \(B = 1.5 \times 10^{-11} \mathrm{T}\) To find the magnitude of the electric field, E, we first differentiate \(B\) with respect to time, and then multiply by the negative sign: \(\frac{d\Phi_B}{dt} = B_0 \cdot 2\pi f \cdot \cos(2\pi f t)\) Plugging in the given values and solving for E: \(E = - \frac{d\Phi_B}{dt} = -(2.5 \times 10^{-11} \mathrm{T}) \cdot (2 \pi \times 3 \times 10^6 \mathrm{s^{-1}}) \cdot \cos(2\pi (3 \times 10^6 \mathrm{s^{-1}})(0))\) \(E = 4.713 \times 10^{-3} \mathrm{V/m}\) The direction of the electric field will be in the \(+x\) direction due to the right-hand rule, given the magnetic field in the \(+z\) direction and the wave traveling in the \(-y\) direction. So, the magnitude of the electric field at \(y=0\) and \(t=0\) is \(4.713 \times 10^{-3} \mathrm{V/m}\), and the direction is in the \(+x\)-direction.