Part 3. The strength of the horizontal component of the Earth's magnetic field According to equation (1) the field at the center of the loop is directly proportional to the product N-I. Now you will test the complete equation by using it to calculate the horizontal component of the Earth's magnetic eld SEEM and compare it with the value from part 2. You will do that as follows. Using a convenient point on the line of best fit from near the middle of your graph tan' vs. I find the value of the current and by equation (1) calculate 31.00;: for that value of the current (mt: do your calculation for the current when rand = 1).Assume that the radius R=20 cm for circular loop. Compute the horizontal component of Earth's magnetic field BEN, using equation (2) for given value of rand and calculated value of BLDG? (notice if rand = 1, then BM = 315015.). Compare your calculated value of the earth's magnetic field Bmh with the average value from part 2