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Old Feb 8th 2011, 12:44 PM   #1
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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
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Old Feb 8th 2011, 04:29 PM   #2
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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)
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Old Feb 8th 2011, 05:14 PM   #3
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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?
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Old Feb 8th 2011, 09:04 PM   #4
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D all of the above,a/c makes transformer use possible..
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Old Feb 9th 2011, 12:36 AM   #5
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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
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Old Feb 11th 2011, 12:13 PM   #6
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Thanks for your help
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