The scalar and vector potentials generated by a wire loop of radius b with alter

Question

The scalar and vector potentials generated by a wire loop of radius b with alternating cur- rent I(t) = I_0 cos(omega t) (i.e., a magnetic dipole antenna) in the a centered at the origin v-plane are given by (given in spherical coordinates): V(r, theta, phi; t) = 0 A(r, theta, phi; t) = - mu_0b^2I_0 omega/4c(sin theta/r) sin (omega(t - r/c)) phi solution assumes the typical radiation approximation r >> c/omega >> b. Derive the at distances far away from the wire loop. Your expression s consist of two of the two terms can be ignored and why? Briefly explain your answer.

Explanation / Answer

Solution :-
We know that

C (r, theta, phi, t) = 1/ 2*pi*1 * Sum (Sum Y (L,m) (Thta, phi)

And given V (r, theta, phi, t) = 0

and A (r, theta, phi, t) = -uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi)

Magntiude at distance far away from the wire loop = 1/ 2*pi*1 * Sum (Sum Y (L,m) (Thta, phi) /-uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi) + V* -uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi)

= 1/ 2*pi*1 * Sum (Sum Y (L,m) (Thta, phi) /-uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi) + 0* -uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi)

= 1/ 2*pi*1 * Sum (Sum Y (L,m) (Thta, phi) /-uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi) +0 (Because V = 0)

= 1/ 2*pi*1 * Sum (Sum Y (L,m) (Thta, phi) /-uob^2Iow/4c (Sin(theta)/r) Sin (w(t-r/c) phi)

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