What is the induced emf in a radio antenna receiving an AM radio signal?

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SUMMARY

The induced electromotive force (emf) in a radio antenna receiving an AM signal at 800 kHz is calculated using the formula emf = -Nd[flux]/dt. The antenna consists of an inductor with 120 turns and a radius of 0.6 cm, resulting in an induced emf of -0.682cos(5.03*10^{-6}t) V. The magnetic field is given by \vec{B}=1.0*10^{-5}sin(2\pi ft)\vec{k}, and the area is calculated as A = πr². The integration of the magnetic field over the area leads to the correct flux calculation, confirming the solution.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with calculus, specifically integration and differentiation
  • Knowledge of the formula for induced emf
  • Basic concepts of radio frequency signals and their properties
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  • Learn about the calculation of magnetic flux and its applications
  • Explore the relationship between frequency and induced emf in antennas
  • Investigate the effects of antenna design on signal reception
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royman19
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Homework Statement



A radio antenna that receives an AM radio station that emits at 800 kHz is constituted of an inductor of 120 turns with a radius of 0.6cm. The inductor is home to an iduced emf du to a magnetic field oscillating from the radio wave. The magnetic field is \vec{B}=1.0*10^{-5}sin(2\pift)\vec{k}. What is the induced emf in the inductor? We suppose that the magnetic field is oriented according to the inductor's axe.

Homework Equations


emf = -Nd[flux]/dt

d[flux] = int[B*dA]

The Attempt at a Solution



I tried to integrate and find the flux, but I am not that great at calculus, but then what I really don't get is that I am going to have to derive to find the emf. It seems like I am going in circles. By the way, the final answer is -0.682cos(5.03*10^{-6}t) V.

Thanks for the help!
 
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I'm checking it right now, but seems to me that B here doesn't vary with area or location or anything

I'm assuming that the d in front of [flux] is a typo, as is the 2^pi(should be 2pi)

So you're integrating B*dA...>_> if B doesn't depend on the area, it just comes out of the integral and you get B*A, which is pi*r^2

then you take the time derivative of THAT(which most certainly depends on time)times -N

Your concern would be right except you're integrating over dA, then differentiating with respect to time

Edit: This is correct, and I checked it on windows calculator so it took like 6 start overs when I clicked the wrong thing :(
 
Last edited:
Thank you very much for the help, it works now. I was just complicating the problem. For some odd reason I was integrating the function for the magnetic field in the flux calculation, but it all works now. Thank you!
 
Last edited:

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