GR8677 #53



Alternate Solutions 
flyboy621 20101114 21:12:43  I don't know if this will help anybody, but...
You can imagine standing at the origin, holding one end of a rope, the other end of which is at point P. Then you start waving the end of the rope back and forth along the xaxis. The waves in the rope will be oriented in the xyplane and be maximum when P is on the yaxis, i.e. theta is 90 degrees.
The analogy works because EM waves are transverse, just like waves in a string.   Herminso 20090922 13:56:33  For a electric dipole we have:
and
Thus the oscilation of the electric field is in the xyplane and the maximum is at , just the yaxes.
  ee7klt 20051111 04:54:58  hi,
since oscillations happen only in the xy plane, the Efield vector thus cannot have components in the zdirection. This eliminates (A) and (B).
The minimum occurs when you're looking directly down the xaxis i.e. (all you see is the infinitesimal tip of the field vector which doesn't look like it'll generate any waves coming towards you)  this eliminates (D). From here, I guess you'll need to remember that the field goes as to narrow it further down to (C).  

Comments 
santo35 20140731 03:59:21  Isnt the electric field varying at all points "P" ? (since it is an oscillation ?
And if so are we considering the maximum value of variation here ?   santo35 20140731 03:58:36  Isnt the electric field varying at all points "P" ? (since it is an oscillation ?
And if so are we considering the maximum value of variation here ?   santo35 20140731 03:56:44  isnt that field varying at all points P ? rnrnOr is it that we are considering only the maximum value of the variation at all points P and comparing them ?   849444209 20131016 07:48:56    natec 20130818 17:46:35  Say point P were in the XZ plane and was the angle BELOW the xaxis. Would the electric field now be in the XZ plane with maximum magnitude at ? I can't see why not because of the symmetry of the arrangement.   SillyMan 20130619 20:17:23  The solution given above is wrong. Purely qualitatively, by Ampere's Law, B field will be in the +/ Z direction at point P. The energy flux vector S points outward (obviously). Thus E is in the XY plane (Orthogonal Triad). At 90 degrees, one "sees" the highest acceleration, which means that the power flowing past P is the highest at theta = 90 degrees. The answer was implied by Maxwell's equations and the Larmor result.   flyboy621 20101114 21:12:43  I don't know if this will help anybody, but...
You can imagine standing at the origin, holding one end of a rope, the other end of which is at point P. Then you start waving the end of the rope back and forth along the xaxis. The waves in the rope will be oriented in the xyplane and be maximum when P is on the yaxis, i.e. theta is 90 degrees.
The analogy works because EM waves are transverse, just like waves in a string.
deafmutemouse 20110920 14:10:54 
This analogy is awesome! Thanks!

OptimusPrime 20170408 04:47:54 
This is really helpful! The point of fastest velocity of the rope shaking happens when crossing the yaxis, which is maximum E amplitude in this case.

  faith 20101111 23:14:24  here's another way to look at it.
cross product! v=ExB since particle is moving along the x direction, E field max should only be along y.
faith 20101111 23:18:35 
yikes.. sorry, this solution is wrong. i had it by luck.

  wittensdog 20090928 21:43:42  The one thing I'm seeing over and over again is that every physics GRE problem has a quick and simple way to do it. Sometimes it's a calculation trick, sometimes it's just knowing something by heart.
In this case, I strongly recommend remembering two basic facts which have a good chance of coming up on any test:
1.) an oscillating charge never radiates in the direction of its oscillation axis
2.) the polarization of an oscillation charge is parallel to the oscillation axis
Don't bother wondering why (at least not for the sake of the GRE), just memorize that. If you use those two pieces of information, you immediately see that the Efield should be in the xy plane from the restriction on the polarization, and also that the maximum field strength should be at 90 degrees, since along the x axis it has no magnitude, and increases as you move away from the x axis.
So far, in all of my studying, I've never come along a GRE problem that required a formula with more than 3 or 4 terms in it. Maybe that's a slight exaggeration, but you get the point.
apr2010 20100409 09:22:19 
Does not the dipol radiate also in the zdirection? This looks like a trap again, as the Point P lies only in the xyplane, meaning A and B are not even available.

  Herminso 20090922 13:56:33  For a electric dipole we have:
and
Thus the oscilation of the electric field is in the xyplane and the maximum is at , just the yaxes.
Herminso 20090922 14:22:02 
Where is measured from the positive xaxes on the xyplane.

  a19grey2 20081103 22:14:06  I recommend just learning what the picture of an oscillating dipole looks like. rnhttp://en.wikipedia.org/wiki/DipolernrnIt'll help you later in your physics life anyway...
physicsisgod 20081105 20:32:44 
Here's a cool video of it:
http://www.vis.unistuttgart.de/ufac/dipole/

  jw111 20080914 11:09:42  You see the maximum OSCILATION of charge when you stand on y axis, and minimum oscilation when you stand on x axis.
=> max field take place on y axis
the ocilation is on xy plane, the field line wave mainly on xy plane.
=>oscilating field on xy plane   ee7klt 20051111 04:54:58  hi,
since oscillations happen only in the xy plane, the Efield vector thus cannot have components in the zdirection. This eliminates (A) and (B).
The minimum occurs when you're looking directly down the xaxis i.e. (all you see is the infinitesimal tip of the field vector which doesn't look like it'll generate any waves coming towards you)  this eliminates (D). From here, I guess you'll need to remember that the field goes as to narrow it further down to (C).
Simplicio 20090331 17:13:47 
It's oscillating along the xaxis so you can't just say "it is in the xy plane". It might as well been the xz plane ...

  physicsDen 20051109 22:04:58  is it me, or is this a poor attempt at a solution?
alpha 20051109 22:15:19 
it's a quick and dirty way to arrive at the right answer.

mpdude8 20120419 14:19:49 
Definitely a poor attempt at a solution, but a great solution for the sake of the GRE.
You do not have the time to use any formulas on a problem like this. It really has to be 100% intuition in order for you to finish. You don't have 34 minutes to think of a relevant formula, you just have to see the problem and know what to look for instinctively.
Thus, many of the solutions to these problems on this site are not rigorous whatsoever, but the GRE couldn't care less about rigor.

 




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