Question

vs. I, which you plotted in the lab magnetic Using the slope and uncertainty from the logger pro graph of tan(0) fields," calculate the value of Bearth (in Teslas) with the error assuming that the radius of the ring is R-20cm and the number of turns is N- 5 Linear Fit for: Data Set I tan(theta) tan(theta) my+b m (Slope): 0.7482 +0.04211 35 3 2.14S b (YIntercept): -0.07427 -0.08808 Correlation: 0.9906 RMSE: 0.1365 70 #0 12 13 14 18 17 20 21 24 25 30

Answer 1

From graph Tan theta = m I + Tan that_0 m I = Tan theta - tan theta_0 given r = 20 cm = 0.2 m N = 5 we know B = B_H tan theta mu_0 N I/2 r = B_H (tan theta - tan theta_0) 1/B_H = 2r/mu_0 N [tan theta /I - tan theta_0/I] 1/B_H = 2 times 0.2/4 times 3.14 times 10^-7 times 5 [0.7482 - (-0.0743)] 1/B_H = 2 times 0.2 times 0.8225/4 times 3.14 times 5 times 10^-7 B_H = 4 times 3.14 times 5 times 10^-7/2 times 0.2 times 0.8225 = 190.88 times 10^-7 B_H = 0.191 times 10^-4 T or B_H = 0.191 G or B_H = 0.191 times 10^-4 T = 19.1 mu T therefore Horizontal component of earth's magnetic field is = 19.1 mu T = 0.19G