## Radiating Current Element

Figure 9.5 shows a short current element at the origin. The element of a length dz along the z-axis carries an alternating sinusoidal current 10, which is constant along the element. This kind of current element is also called

Figure 9.5 shows a short current element at the origin. The element of a length dz along the z-axis carries an alternating sinusoidal current 10, which is constant along the element. This kind of current element is also called

Figure 9.5 Radiating current element.

the Hertz dipole. In his experiments, Heinrich Hertz used end-loaded dipoles. Because of the capacitive loadings, currents could flow even at the ends of the dipole making a nearly constant current distribution possible. As explained in Chapter 2, a fluctuating current produces electromagnetic waves: The current produces a changing magnetic field, the changing magnetic field produces a changing electric field, the changing electric field produces a changing magnetic field, and so on.

The volume integral of the current density is fy J dV = Iq dzuz in the case of a current element. Therefore, the vector potential at a point P (r, 0, <) is

The components of the vector potential in the spherical coordinate system are A r = A z cos 0, Aq = —Az sin 0, and A< = 0. Equations (9.8) and (9.9) give the components of the fields:

where V is the wave impedance in free space.

Those components of the field having a 1/r2 or 1/r3 dependence dominate at small distances but become negligible at larger distances. Far away from the element, the fields are jM/I0dz _jkr . a ,Q

Other components are negligible. The electric and magnetic fields are in phase and perpendicular to each other, just like in the case of a plane wave.

The power radiated by the current element is calculated by integrating the Poynting vector S = (1/2) Re (E X H*) over a sphere surrounding the element. If the radius of the sphere, r, is much larger than dz, S is perpendicular to the surface of the sphere and the outflowing power per unit area is 5 = | S | = (1/2) | Ee H^ |. The surface element for integration is selected as shown in Figure 9.6. The power radiated is

The radiation resistance of the current element is obtained by equating the radiated power with P = (1/2) RrI0 :

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