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Magnetic Flux Location

  • Thread starter Octoshark
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  • #1
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Homework Statement



At a certain location, the Earth's magnetic field has a magnitude of 5.6 x 10^−5 T and points in a direction that is 65° below the horizontal. Find the magnitude of the magnetic flux through the top of a desk at this location that measures 110 cm by 62 cm

Homework Equations



flux = BAcos(theta)

The Attempt at a Solution



flux = (5.6x10^-5)(.682)(cos65)

flux = 1.614 x 10^-5 Wb (which is incorrect). Where did I go wrong?
 

Answers and Replies

  • #2
kuruman
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How is angle theta in your expression for the flux defined and how is that theta angle related to the 65o that is given?
 
  • #3
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I'm not sure what you mean but all my teacher said is you measure the angle by saying "Angle of the axis of the object against B"
 
  • #4
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Got the answer.

I just messed around with some things and tried sin(65) instead of cos(65) and got the right answer of 3.46x10^-5 Wb
 
  • #5
kuruman
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Messing around with "things" does not promote understanding of what is going on. You will not have enough time on a test to mess around, especially if you don't know what the correct answer is. Here is what is going on.

Angle θ in Φ = BA cos(θ) is the angle that the B field makes with respect to the perpendicular (normal) direction to the loop. You are given that the field makes an angle of 65o with respect to the plane of the loop. Since the normal to the loop is always at 90o with respect to the plane of the loop, this means that the angle that the field with respect to the normal, i.e. θ, is θ = 90o - 65o = 25o. Now sin(65o=cos(25o) and that's why you got the right answer.
 

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