1-The total magnetic field intensity at the location of Edmonton was 57652 nT on

Question

1-The total magnetic field intensity at the location of Edmonton was 57652 nT on March 20, 2012, when I first wrote this question. On Feb. 19, 2013, it was 57569 nT. On January 6, 2014 it is 57,489 nT. Random noise, or is it getting weaker like in the video? Anyway, at Edmonton 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 1,000 cm2 and is horizontal. Given the values of the magnetic field I listed for Jan.06, 2014, what was the magnetic flux through the desk? Use SI units.

answer with 5 sig fig.

2-This is an MRI question. The magnetic field region of an MRI is typically on the order of 1 meter long, in order to keep the machine at a reasonable size for a human being. A given MRI may be able to reach fields up to several Tesla using superconducting wire that is 1 mm in diameter, which can typically carry a current of 1000 Amps. In order to reach such high fields, there is no choice but to form the wire into layered (or

Explanation / Answer

(1)Treating the lightening bolt as long straight conduction .. and applying
B = ??.i / 2?d for the field at a distance d

B(l) = (4?^-7) x 250,000A / (2? x 10m) .. B(l) = 5.0^-3 T

B(l) = n x B(e) .. .. n = B(l) / B(e) = 5.0^-3 / 57.50^-6 .. ?n = 87 (? 90)

(2) The mag force on one bolt due to the field from the other bolt is given by ..
F = BiL .. with the force per m, F/L = B x i

F/L = 5.0^-3 T x 250,000A .. .. ?F/L = 1250 N/m

?=              = BAcos?

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

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

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