N is the total number of turns and r the radius of the circle inside the toroid's core.
Derivation
- Take an Ampère loop of radius r inside the core.
- It encloses the current NI, and B is the same all round it: B × 2πr = μ₀NI.
- B = μ₀NI/2πr. Outside the core the enclosed current is zero, so B = 0.
Points to remember
- Writing n = N/2πr gives B = μ₀nI, the same as a solenoid.
All field formulas
| Field due to | Formula |
|---|---|
| Electric field due to a point charge | E = kq/r² = q/(4πε₀r²) |
| Electric field due to an electric dipole on its axis | E = 2kp/r³ |
| Electric field due to an electric dipole on its equatorial line | E = kp/r³ |
| Electric field due to a charged ring on its axis | E = kQx/(x² + R²)^(3/2) |
| Electric field due to an infinite line charge | E = λ/(2πε₀r) = 2kλ/r |
| Electric field due to an infinite charged sheet | E = σ/(2ε₀) |
| Electric field due to a charged conducting sphere (or spherical shell) | E = kQ/r² outside, 0 inside |
| Electric field due to a uniformly charged non-conducting sphere | E = kQr/R³ inside, kQ/r² outside |
| Magnetic field due to a long straight wire | B = μ₀I/(2πr) |
| Magnetic field due to a finite straight wire | B = (μ₀I/4πd)(sin α + sin β) |
| Magnetic field due to a circular loop at its centre | B = μ₀I/(2R) |
| Magnetic field due to a circular loop on its axis | B = μ₀IR²/(2(R² + x²)^(3/2)) |
| Magnetic field due to a long solenoid | B = μ₀nI |
| Magnetic field due to a toroid | B = μ₀NI/(2πr) |
Questions students ask
What is the magnetic field due to a toroid?
B = μ₀NI/(2πr). N is the total number of turns and r the radius of the circle inside the toroid's core.
How do you derive the magnetic field due to a toroid?
Take an Ampère loop of radius r inside the core. It encloses the current NI, and B is the same all round it: B × 2πr = μ₀NI. B = μ₀NI/2πr. Outside the core the enclosed current is zero, so B = 0.