Field at the Center of Arcs and Loops
A circular arc of radius R and angle phi radians gives B = mu_0 I phi/(4pi R) at its center. A full N-turn circle gives B = mu_0 N I/(2R).
Why this shows up in the exam
Composite arc-and-line conductors · Semicircular and sector coils · Comparing concentric loop contributions
Learn the idea
At a circle's center, a current arc contributes in proportion to its subtended angle. Every element of a circular arc sends field in the same axial direction at the center, so the total scales directly with the angle covered. Radial connector pieces contribute zero there.
🧠 Memory hook: At the center: angle over radius, with radial pieces silent.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- B_arc = mu₀ I phi/(4 pi R) — center field of an arc subtending phi radians
- B_circle = mu₀ N I/(2 R) — center field of an N-turn circular loop
- dB_radial_at_center = 0 — radial segment has dl parallel to r and contributes no field at the center
How to approach it
- 1Split the conductor into arcs and straight pieces
- 2Use radians and right-hand-rule signs
- 3Add signed axial contributions
Common slip-ups that cost marks
- •Using degrees directly in the arc formula
- •Including radial connector fields at the center
- •Adding opposite current senses with the same sign
🌟 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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