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 Electricity and Magnetism Electricity and Magnetism Physics Help Forum Feb 8th 2011, 12:44 PM #1 Junior Member   Join Date: Feb 2011 Posts: 7 Induction in a loop of wire Hi there the problem given is: A circular loop of wire can be used as a radio antenna. If a 22.0-cm-diameter antenna is located 2.20km away from a 95.0-MHz source with a total power of 55.0kW, what is the maximum emf induced in the loop? (Assume that the plane of the antenna loop is perpendicular to the direction of the radiation's magnetic field and that the source radiates uniformly in all directions.) As far as I have gotten in solving the problem: it seems that this problem involves Faraday's law: EMF= d(phi)/dt so as loop as perpendicular to propogation, phi is simply B*A. A=(0.11)^2*3.14 ---> as per the values given in the problem.. to find B: we know intensity, I: 55000/(22000^2*4*3.14) = 9.05*10^-4 W/m^2 also we know I = EmaxBmax/2(mu_0) so can find Bmax from this: => Bmax = 2.76(10^-9) so (phi) = Bmax*A=1.048(10^-10) so to find d(phi)/dt is it simply a matter of dividing the above value by the frequency that was told in the question (95MHz)?........ or the period maybe??? But this apparantly is not the correct answer..... Be very grateful if someone could inform me of whatever mistakes I am making... Andrew   Feb 8th 2011, 04:29 PM #2 Member   Join Date: Jul 2009 Posts: 72 you are using 22km instead of the given 2,2km but the main error i think is you use phi=Bmax * A where it should be phi=B * A, where B=Bmax sin (omega*t) (the wave equation)   Feb 8th 2011, 05:14 PM #3 Junior Member   Join Date: Feb 2011 Posts: 7 Thanks for your reply The 22km was indeed a typo in my post. Would you mind expanding a little on your explanation of using B=Bmax sin (omega*t)? The reason I used Bmax is because the question asked for the maximum EMF induced but that is not correct...? with this alteration you made, should I then take the average value of sin for all omega*t and then the rest of my solution is correct?   Feb 8th 2011, 09:04 PM #4 Banned   Join Date: Feb 2011 Posts: 5 D all of the above,a/c makes transformer use possible..   Feb 9th 2011, 12:36 AM #5 Member   Join Date: Jul 2009 Posts: 72 emf=d(phi)/dt just because phi is maximum doesn't mean d(phi)/dt will be. you dont get the average of Bmax * sin(omega * t), you calculate d(Bmax * sin(omega * t) * A)/dt you need to derive sin(omega * t) in order to t and then you see what is the maximum of that   Feb 11th 2011, 12:13 PM #6 Junior Member   Join Date: Feb 2011 Posts: 7 Thanks for your help  Tags induction, loop, wire Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post antonravik Kinematics and Dynamics 8 Nov 9th 2012 09:49 PM fogo57 Advanced Electricity and Magnetism 2 Oct 16th 2012 05:06 AM bearfrods Electricity and Magnetism 0 Feb 26th 2012 10:06 PM rcmango Electricity and Magnetism 2 Jul 31st 2011 11:22 PM pinkprincess08 Advanced Electricity and Magnetism 1 May 2nd 2009 07:52 PM