I is the current and R the radius. For N turns, multiply by N.
Derivation
- Every element dl is at distance R and perpendicular to the line joining it to the centre, so sin θ = 1.
- dB = (μ₀/4π) I dl/R², all in the same direction.
- Add round the loop (total length 2πR): B = (μ₀/4π) I × 2πR/R² = μ₀I/2R.
Points to remember
- An arc subtending angle θ (radians) gives B = μ₀Iθ/4πR.
- A semicircle gives μ₀I/4R.
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 circular loop at its centre?
B = μ₀I/(2R). I is the current and R the radius. For N turns, multiply by N.
How do you derive the magnetic field due to a circular loop at its centre?
Every element dl is at distance R and perpendicular to the line joining it to the centre, so sin θ = 1. dB = (μ₀/4π) I dl/R², all in the same direction. Add round the loop (total length 2πR): B = (μ₀/4π) I × 2πR/R² = μ₀I/2R.