Problem

A Hertzian dipole is a short conducting wire carrying an approximately constant current ov...

A Hertzian dipole is a short conducting wire carrying an approximately constant current over its length l. If such a dipole is placed along the z-axis with its midpoint at the origin, and if the current flowing through it is i(t) = Io cos ωt. find the following:

(a) The retarded vector potential Ã (R. θ, ɸ) at an observation point Q(R. θ, ɸ) in a spherical coordinate system.

(b) The magnetic field phasoi
(R. θ, ɸ).

Assume I to be sufficiently small so that the observation point is approximately equidistant to all points on the dipole; that is. assume R' R.

Step-by-Step Solution

Solution 1

Step #1 of 3

A Hertzian dipole short conducting wire of length / placed along the $z$-axis carrying a constant current / which is given by,

$$ i(t)=I_{0} \cos \omega t $$

(a) Evaluate the phasor form of the given current which is,

$$ \tilde{I}=I_{0} $$

Integrate the volume as a double integral over the wire cross section and a single integral over its length as follows:

$$ \bar{A}=\frac{\mu}{4 \pi} \int_{-V 2}^{v 2} \iint \frac{\bar{J}\left(\mathbb{R}_{i}\right) e^{-j v}}{R^{\prime}} d z d z $$

Here, $s$ denotes the cross-section of the wire.

Substitute $\tilde{\mathbf{J}}=\hat{\mathbf{z}} \frac{I_{0}}{s}$ in equation (1).

It is given that the conducting wire is so thin, so the distance $R^{\prime}$ in the denominator of equation (1) is not a function of $x$ or $y$.

Further simplification of equation (1) yields,

$$ \begin{aligned} \tilde{\mathbf{A}} &=\hat{\mathbf{z}} \frac{\mu I_{0}}{4 \pi} \int_{-1 / 2}^{n / 2} \frac{e^{-j k^{\prime}}}{R^{\prime}} d z \\ & \approx \hat{\mathbf{z}} \frac{\mu I_{0}}{4 \pi} \int_{-/ / 2}^{n_{2}} \frac{e^{-j t R}}{R} d z \\ &=\hat{\mathbf{z}} \frac{\mu I_{0} I}{4 \pi R} e^{-j k R} \end{aligned} $$

Step #2 of 3

We have to express $\overline{\mathbf{A}}$ in terms of spherical co-ordinate system.

Therefore, $\hat{z}$ in spherical co-ordinate system is given by,

$$ \hat{\mathbf{z}}=\hat{\mathbf{R}} \cos \theta-\hat{\theta} \sin \theta $$

Substitute $\hat{z}$ and we get,

$$ \overline{\mathbf{A}}=(\hat{\mathbf{R}} \cos \theta-\hat{\theta} \sin \theta) \frac{\mu I_{0} I}{4 \pi R} e^{-j k} $$

Therefore, the retarded vector potential $\overline{\mathbf{A}}(R, \theta, \phi)$ is,

$(\hat{\mathbf{R}} \cos \theta-\hat{\theta} \sin \theta) \frac{\mu I_{0} l}{4 \pi R} e^{-\beta t R}$

Step #3 of 3

(b) Substitute the value of $\tilde{\mathbf{A}}$ in the Faraday's law and evaluate the magnetic field phasor $\overline{\mathrm{H}}(R, \theta, \phi)$ as follows:

$$ \tilde{\mathbf{H}}=\frac{1}{\mu}(\nabla \times \tilde{\mathbf{A}}) $$

$$ =\frac{1}{\mu}\left|\begin{array}{ccc} \hat{\mathbf{R}} & \hat{\theta} & \hat{\phi} \\ \frac{\partial}{\partial R} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ \frac{\mu I_{0} l \cos \theta}{4 \pi R} e^{-j R} & \frac{\mu I_{0} l \sin \theta}{4 \pi R} e^{-j k R} & 0 \end{array}\right| $$

$$ \begin{aligned} &=\hat{\phi}\left(\frac{I_{0} I}{4 \pi}\right)\left(\frac{\partial}{\partial R}\left(\sin \theta \frac{e^{-j R}}{R}\right)-\frac{\partial}{\partial \theta}\left(\cos \theta \frac{e^{-j \omega t}}{R}\right)\right) \\ &=\hat{\phi} \frac{I_{0} l \sin \theta e^{-j k R}}{4 \pi R}\left(j k+\frac{1}{R}\right) \end{aligned} $$

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