Can I get the answers for these physic: Theory of Electricity and Magnetism II problems?

Textbook: http://www-pnp.physics.ox.ac.uk/~gwenlan/teaching/EM/introduction-to-electrodynamics-d-j-griffiths-4ed.pdf

Figure 7.28 A square loop, side a, resistance R, lies a distance s from an infinite straight wire that carries current / (Fig. 7.28). Now someone cuts the wire, so that I drops to zero. In what direction does the induced current in the square loop flow, and what total charge passes a given point in the loop during the time this current flows? If you don't like the scissors model. turn the current down gradually: ) (O =an)l, for0 1/a. N PAEquation 9.36 describes the most general linearly polarized wave on a string. inear (or \"plane\") polarization (so called because the displacement is parallel to a fixed vector n) results from the combination of horizontally and vertically polarized waves of the same phase (Eq. 9.39). If the two components are of equal amplitude, but out of phase by 90 (say, 8y =0, 8, = 90), the result is a circularly polarized wave. In that case: (a) At a fixed point z, show that the string moves in a circle about the z axis. Does it go clockwise or counterclockwise, as you look down the axis toward the origin? How would you construct a wave circling the other way? (In optics, the clockwise case is called right circular polarization, and the counterclockwise, left circular polarization.) (b) Sketch the string at time r = 0. (c) How would you shake the string in order to produce a circularly polarized wave? X (a) Vertical polarization {b) Horizontal polarization {c) Polarization veclor f(z,1) = Ae'a @ j, (9.36) Problem 3 Show that the mode TEoo cannot occur in a rectangular wave guide. [Hint: In this case w/c = k, so Eqs. 9.180 are indeterminate, and you must go back to 9.179. Show that Bz is a constant, and hence-applying Faraday's law in integral form to a cross section-that Bz = 0, so this would be a TEM mode.] [Also remember that in a simply connected guide, TEM modes do not exist.] a By aBx iw aEy dEx Ez, (i) = iwBz, (iv) ax ay ax ay aBZ iw aEz ikEy = iwBx, (v) - ik By = - 5Ex (9.179) (ii) ay ay iw aEz aBz =iwBy, (vi) ikBx - = (iii) ikEx ax 2 Ey. ax aEz aBz (i) Ex = k . + w- (w/c) 2 - k2 ax ay i aEz a Bz (ii) Ey =(/c)2 - k2 ay ax (9.180) i a Bz w dEz (iii) Bx = k. (w/c)2 - k2 c2 ay a Bz waEz (iv) By = (w/c)2 - k2 ay + 2 ax