Field Inside a Current-Carrying Cylinder
For a long solid wire of radius R carrying uniformly distributed current I, B(r) = mu_0 I r/(2pi R^2) for r <= R and B(r) = mu_0 I/(2pi r) for r >= R.
Why this shows up in the exam
Interpreting B-versus-r graphs · Finding half-maximum locations · Analyzing uniform current density
Learn the idea
Inside a uniformly conducting wire, enclosed current grows with area, so magnetic field grows linearly with radius. Near the axis, a small Amperian circle encloses only a small fraction of the total current. At the surface the field is maximum, and outside it falls as one over distance.
🧠 Memory hook: Inside rises with r; outside falls with one over r.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- I_enclosed = I r²/R² — uniform-current fraction enclosed at radius r inside the wire
- B_inside = mu₀ I r/(2 pi R²) — linear interior field
- B_outside = mu₀ I/(2 pi r) — exterior field
- B_max = mu₀ I/(2 pi R) — continuous maximum at the surface
How to approach it
- 1Write enclosed current from area ratio
- 2Apply Ampere's law inside and outside separately
- 3Check continuity and the surface maximum
Common slip-ups that cost marks
- •Using the full current at every interior radius
- •Making B discontinuous at the surface
- •Assuming linear behavior outside
🌟 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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