Magnetizing Field of Solenoids and Toroids
For a long solenoid H = nI and B = mu H. For a toroid H = NI/(2pi r) locally. In a linear core, B_core/B_air = mu_r and the fractional increase is approximately chi_m.
Why this shows up in the exam
Core-filled solenoids · Toroid field changes · Finding current or turns from magnetic intensity
Learn the idea
Free current fixes H in an ideal long solenoid or toroid, while the core permeability fixes B. The winding supplies the magnetizing effort H. Inserting a linear core adds material response, multiplying the resulting B by its relative permeability without changing the free-current value of H for the same ideal geometry.
🧠 Memory hook: Winding sets H; material scales B.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- H_solenoid = n I = N I/L — magnetic field intensity inside a long solenoid
- B = mu₀ mu_r n I — flux density with a linear core
- H_toroid = N I/(2 pi r) — ideal toroid magnetizing field at radius r
- (B_core-B_air)/B_air = chi_m — fractional increase for a linear material
How to approach it
- 1Compute n = N/L in SI units
- 2Find H from free current first
- 3Use mu_r or chi only when converting H to B or M
Common slip-ups that cost marks
- •Using B = mu₀ nI after inserting a magnetic core
- •Multiplying H by mu_r when free current and geometry are unchanged
- •Forgetting to convert turns per centimetre to turns per metre
🌟 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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