Earth's Magnetic Field, Dip, and Compass Response
If total field B makes dip angle delta below the horizontal, B_H = B cos delta and B_V = B sin delta. A horizontal magnetic needle has T = 2pi sqrt(I_rot/(m B_H)) for small oscillations.
Why this shows up in the exam
Compass oscillation comparisons · Finding total Earth field from dip · Magnetometer measurements
Learn the idea
A horizontal compass responds to the horizontal component of Earth's field, not the total field. Earth's field is tilted. A freely dipping needle follows the total field, but a needle constrained to a horizontal plane feels only the horizontal component, which controls its restoring torque and period.
🧠 Memory hook: Horizontal needle sees B cos dip.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- B_H = B cos(delta) — horizontal component of Earth's field
- B_V = B sin(delta) — vertical component of Earth's field
- f² proportional to B cos(delta) — frequency comparison for the same horizontal compass needle
How to approach it
- 1Draw total field and resolve it
- 2Use only B_H for horizontal-needle torque
- 3Form a ratio with f squared or T squared
Common slip-ups that cost marks
- •Using B sin delta for the horizontal component
- •Comparing frequencies without squaring them
- •Ignoring a change in dip angle between locations
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.
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