Electromagnetic Waves and Induction

AI Thread Summary
The discussion focuses on calculating the magnetic flux through a loop positioned around a solenoid. The magnetic flux is determined using the formula Φ = AB cos(θ), where A is the area of the loop, B is the magnetic field strength, and θ is the angle between the magnetic field and the normal to the loop. When the loop is perpendicular to the solenoid, the magnetic flux is calculated to be 6.3×10−5 Wb. For a 60-degree tilt, the participants discuss using the same formula but express confusion over the correct application of the cosine function and area calculations, leading to discrepancies in their results. The conversation highlights the importance of understanding the relationship between the solenoid's magnetic field and the loop's orientation in calculating magnetic flux.
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Homework Statement



The 2.0-cm-diameter solenoid in the figure passes through the center of a 6.0-cm-diameter loop. The magnetic field inside the solenoid is 0.20 T.
What is the magnetic flux through the loop when it is perpendicular to the solenoid?
Φ1 = Wb

What is the magnetic flux through the loop when it is tilted at a 60 angle?
Φ2 = Wb

http://i63.photobucket.com/albums/h148/jlmessick88/jfk_Figure_25_P08.jpg

Homework Equations



Φ = AB cos()

The Attempt at a Solution


? No idea :(

Thanks
 
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You have the equation for calculating Φ. You have B. You have A. You have θ. What seems to be the trouble?
 
getting confused with what i should be doing with the solenoid and the loop...wouldn't one effect the other or am i supposed to be using both diameters
 
The question asks what the magnetic flux through the loop is. The loop does not have a magnetic field, so how could it affect the solenoid?
 
oh! ok that makes sense...
but for when it's tilted 60 degrees... it would be .01^2 * pi = 3.14*10^-4 *.2T = 6.28*10^-5 * cos(60) = 3.14 * 10^-5...but that's not the correct answer...i don't understand why...
 
The question asks what the magnetic flux through the loop is. The loop does not have a magnetic field, so how could it affect the solenoid?

EDIT:

cepheid said:
The question asks what the magnetic flux through the loop is.
 
What is the magnetic flux through the loop when it is perpendicular to the solenoid?
Express your answer using two significant figures.
=6.3×10−5
Correct
.01^2 * pi = 3.14*10^-4
3.14*10^-4 *.2 = 6.28*10^-5
6.28*10^-5 * cos(0) = 6.28*10^-5

so for the second part wouldn't i just use cos(60)??
 
jlmessick88 said:
What is the magnetic flux through the loop when it is perpendicular to the solenoid?
Express your answer using two significant figures.
=6.3×10−5
Correct
.01^2 * pi = 3.14*10^-4
3.14*10^-4 *.2 = 6.28*10^-5
6.28*10^-5 * cos(0) = 6.28*10^-5

so for the second part wouldn't i just use cos(60)??

Yes, you just use flux=BAcos(60). I don't know why you're doing:

.01^2 * pi = 3.14*10^-4
3.14*10^-4 *.2 = 6.28*10^-5
6.28*10^-5 * cos(0) = 6.28*10^-5
 
but when i used BAcos(60), i get 3.14*10^-5, that's wrong...why??
 
  • #10
Not really seeing why the answer is wrong. What does your book say?
 
  • #11
i don't know what the answer is...it's just saying it's wrong...
 
  • #12
Its the same answer for both.. even though I'm sure this is too late now
 
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