What is the magnetic flux through a desk at a certain location on Earth?

[Octoshark]

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?

 

[kuruman]

How is angle theta in your expression for the flux defined and how is that theta angle related to the 65^o that is given?

 

[Octoshark]

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"

 

[Octoshark]

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

 

[kuruman]

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 65^o with respect to the plane of the loop. Since the normal to the loop is always at 90^o 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 = 90^o - 65^o = 25^o. Now sin(65^o=cos(25^o) and that's why you got the right answer.