Field of a Straight Current-Carrying Wire
At perpendicular distance a from a finite straight segment, B = (mu_0 I/(4pi a))(sin theta_1 + sin theta_2), with endpoint angles measured from the perpendicular. The infinite-wire limit is mu_0 I/(2pi a).
Why this shows up in the exam
Finding fields near long power conductors · Evaluating finite leads in composite-wire problems · Comparing field strengths at different distances
Learn the idea
A straight wire creates circular magnetic field lines whose strength depends on distance and endpoint angles. The field wraps around the wire. An infinite wire has the familiar inverse-distance result, while a finite wire contributes less because its endpoints cut off the current path.
🧠 Memory hook: Straight wire fields circle the wire and weaken as one over distance.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- B_finite = (mu₀ I/(4 pi a))(sin(theta₁) + sin(theta₂)) — finite straight-wire field under the stated endpoint-angle convention
- B_infinite = mu₀ I/(2 pi a) — limit for a wire effectively infinite in both directions
How to approach it
- 1Decide finite or effectively infinite
- 2Measure perpendicular distance and endpoint angles
- 3Assign direction and add other segment fields
Common slip-ups that cost marks
- •Using the infinite-wire formula for a visibly finite segment
- •Measuring endpoint angles from the wire instead of the perpendicular
- •Missing the right-hand-rule direction
🌟 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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