Hello again. :) I think I figured out how to do this one, but I needed to google the instrument to learn what they were talking about. ## \\ ## If you define a coordinate system with ## \hat{z} ## upward , ## \hat{y} ## pointing north, and ## \hat{x} ## pointing eastward, you can determine the vector components of the unit vector ## \hat{a} ## for the direction of the magnetic field ## \vec{B}=B_o \hat{a} ##. Call that vector ##\hat{a}=a_x \hat{i}+a_y \hat{j} +a_z \hat{k} ##. The ## a_z ## component is the easiest one to calculate. Since the angle (dip angle) the magnetic field makes with the horizon is 30 degrees, (and this means that ## a_z ## is negative but that isn't important), you can compute the ## a_z ## component. (Note that this vector has unit length along the hypotenuse). ## \\ ## The 45 degrees declination means the magnetic field is in the northeast direction rather than north. Would you agree that ## a_x=a_y ##? You already computed ##a_z ## without computing ## a_x ## or ##a_y ##. Since ## \hat{a} ## is a unit vector, you should now be able to compute ## a_x ## and ## a_y ##. ## \\ ## The next part should get you to the answer: This instrument is set up along the geographic meridian, so that it will apparently only be affected by the ## y ## and ## z ## components of the magnetic field. Essentially the magnetic field of relevance is ## \vec{B}_{relevant}=B_o(a_y \hat{j}+a_z \hat{k}) ##. (The instrument apparently can't respond to the x component when you line it up along the north (y) direction). You should now be able to compute how the instrument will point in response to the portion of the magnetic field that lies in the y-z plane, ignoring the x-component. ## \\ ## (Note: Had the angle been some arbitrary angle ## \theta ## , in general ## \frac{a_x}{a_y}=\tan{\theta} ##. For ## \theta=45 ## degrees, it made it simple). ## \\ ## See also:
https://www.britannica.com/technology/dip-circle They specified geographic meridian here and not magnetic meridian, as in the article. In this case, they apparently lined up the wheel of the instrument, to which a needle is attached, in the y-z plane. The wheel is free to spin in the y-z plane, (with its axle along the x-axis). The compass needle is attached to the wheel, and will point along the magnetic field direction that is in the y-z plane. (It might interest you, the compass needle is a long permanent magnet, and when placed in a magnetic field, the needle aligns with the magnetic field, with the" +" pole of the needle in the direction of the magnetic field, and the "-" pole opposite the magnetic field direction).## \\ ## And yes, I get an answer that is listed. See if you get the same answer. Also, it might be worthwhile to compare this angle, (plug ## \tan^{-1}x ## into the calculator), to the "true" dip angle=the angle they gave you. ## \\ ## Additional question for you: What would be the tangent of the dip angle that this instrument measures if you line up the plane of the wheel so that it now is in the direction of the magnetic meridian? (i.e. you use a compass to line up the wheel to get the "true" dip reading).
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