# Finding magnetic flux

1. Mar 24, 2013

### monnapomona

1. The problem statement, all variables and given/known data
The total magnetic field intensity at the location of Edmonton was 57652 nT on March 20, 2012, when I first wrote this question. Now, on Feb. 19, 2013, it is 57569 nT. The magnetic inclination is +75 degrees (i.e., it points downward at an angle of 75 degrees with respect to the horizontal). This site will tell you the current value of the earth's magnetic field parameters at any location you choose.

Consider the desk you sit at in our class. Let us estimate that it has an area of about 900 cm^2 and is horizontal. Given the values of the magnetic field I listed for Feb. 19, 2013, what was the magnetic flux through the desk? Use SI units.

Note: Calculate and enter your answer with at least 5 significant digits. I had to make the accepted range of values in this answer quite small.

2. Relevant equations

$\Phi$ = BAcos$\vartheta$

3. The attempt at a solution

I used A = 0.09 m^2 and B = 57569 x 10^-9. For the angle, I used 90 deg - 75 deg = 15 deg.

$\Phi$ = (57569 x 10^-9 T)(0.09 m^2)cos(15)
= 0.0000050050 Tm^2

This was wrong in my online assignment. Not sure where I'm going wrong with this... Am I using the wrong angle?

2. Mar 24, 2013

### TSny

Everything looks good, except the answer you give is not quite accurate to 5 significant figures. Also, scientific notation will make it look nicer.

3. Mar 24, 2013

### monnapomona

Oh I see now! Thank you for your help and clarification!