Electromagnetic Waves and Induction

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SUMMARY

The discussion centers on calculating magnetic flux through a loop in relation to a solenoid. The magnetic field inside the solenoid is given as 0.20 T, and the loop has a diameter of 6.0 cm. The magnetic flux when the loop is perpendicular to the solenoid is calculated as 6.28×10-5 Wb. When tilted at a 60-degree angle, the flux calculation involves using the equation Φ = BA cos(θ), leading to confusion regarding the correct application of the cosine function and the area calculation.

PREREQUISITES
  • Understanding of magnetic flux and its calculation
  • Familiarity with the equation Φ = BA cos(θ)
  • Knowledge of solenoids and magnetic fields
  • Basic geometry for calculating the area of a circle
NEXT STEPS
  • Study the principles of electromagnetic induction
  • Learn about the application of the right-hand rule in magnetic fields
  • Explore the effects of angle on magnetic flux calculations
  • Review examples of magnetic flux in different orientations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand magnetic flux calculations in relation to solenoids and loops.

jlmessick88
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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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