d is the perpendicular distance from the wire; α and β are the angles the two ends make with the perpendicular at the point.
Derivation
- Biot–Savart: dB = (μ₀/4π) I dl sin θ/r².
- Put l = d tan φ, so dl = d sec²φ dφ and r = d sec φ.
- dB = (μ₀I/4πd) cos φ dφ; integrate from −α to β: B = (μ₀I/4πd)(sin α + sin β).
Points to remember
- For an infinite wire, α = β = 90° gives μ₀I/2πd.
- At a point beside one end of a semi-infinite wire, B = μ₀I/4πd.
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 finite straight wire?
B = (μ₀I/4πd)(sin α + sin β). d is the perpendicular distance from the wire; α and β are the angles the two ends make with the perpendicular at the point.
How do you derive the magnetic field due to a finite straight wire?
Biot–Savart: dB = (μ₀/4π) I dl sin θ/r². Put l = d tan φ, so dl = d sec²φ dφ and r = d sec φ. dB = (μ₀I/4πd) cos φ dφ; integrate from −α to β: B = (μ₀I/4πd)(sin α + sin β).