On this solution I am curious as for why we start with the trig function (sin ).

Question

On this solution I am curious as for why we start with the trig function (sin ).  I am not looking for a breakdown of the entire problem just the initial setup to allow myself to progress through understanding whats going on.  Thanks and will quickly rate 5 stars!

A horizontal infinitesimal electric dipole of constant current I0 is placed symmetrically about the origin and directed along the x-axis. Derive the far-zone fields radiated by the dipole directivity of the antenna sin Psi = 1 - cos2Psi = 1 - |delta x . ar|2 = 1 - (sin theta cos Phi)2 In far-zone fields E Psi = jetakI0 . le-jkr/4pir.sinPsi = jetaK.I0le-jkr/4pir. 1 - (sin theta cos Phi)2 Hx = jkI0le-jkr/4pir.sin Psi = EPsi/eta U = U0 (1 - sin2theta.cos2Phi) Prod = U0 (1 - sin2theta.cos2Phi).sin theta d theta d Phi = U0.8pi/3 D0 = 4pi.U0/U0.8pi/3 = 3/2 = 1.5

Explanation / Answer

Location of the dipole is in z direction hence we need to take the sin component for the calculation.Hope it helps

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