Question

Write down the difference between displacement current and conduction current. The equation of an electromagnetic wave is \(B_y = 2\times10^{-7}\,\sin(500x + 1.5\times10^{11}t)\) Tesla. Find the equation of the electric field of the wave, the direction of propagation and the wavelength of the wave.

Hint:

Read \(B_0,k,\omega\) from the wave; \(c=\omega/k\), \(E_0=cB_0\); phase \((kx+\omega t)\) means motion along \(-x\); \(\lambda=2\pi/k\).

Answer 1

Step 1: Difference between the two currents.
Conduction current \(I_c\): produced by the actual drift of free charges (electrons) through a conductor; it obeys Ohm's law and exists wherever charges physically move.
Displacement current \(I_d\): produced by a time-varying electric field (changing electric flux), for example in the gap between the plates of a charging capacitor; no real charge crosses the gap. \[ I_d = \varepsilon_0\,\frac{d\Phi_E}{dt} \] Both act as sources of magnetic field, which is why Ampere's law is completed by adding \(I_d\).

Step 2: Read the constants from the wave.
Comparing with \(B_y = B_0\sin(kx+\omega t)\): \(B_0 = 2\times10^{-7}\) T, \(k = 500\ \text{rad/m}\), \(\omega = 1.5\times10^{11}\ \text{rad/s}\).

Step 3: Speed and electric-field amplitude.
\[ c = \frac{\omega}{k} = \frac{1.5\times10^{11}}{500} = 3\times10^{8}\ \text{m/s} \] \[ E_0 = cB_0 = (3\times10^{8})(2\times10^{-7}) = 60\ \text{N/C} \] Step 4: Direction of propagation.
The phase is \((kx+\omega t)\), so the wave travels along the \(-x\) direction. Since \(\vec{B}\) is along \(+y\) and the propagation direction equals \(\hat{E}\times\hat{B}\), the electric field must be along \(+z\) (because \(\hat{z}\times\hat{y}=-\hat{x}\)).

Step 5: Electric-field equation.
\[ E_z = 60\,\sin(500x + 1.5\times10^{11}t)\ \text{N/C} \] Step 6: Wavelength.
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{500} = 1.26\times10^{-2}\ \text{m} \] \[\boxed{E_z = 60\sin(500x+1.5\times10^{11}t)\ \text{N/C},\ \text{along }-x,\ \lambda\approx1.26\times10^{-2}\ \text{m}}\]