Short on time? This page is the recap. For the derivations, the figures and the worked examples, use the full study guide. This topic is in the calculus-based course.
Recap
- Every small piece of current-carrying wire makes a small magnetic field. The Biot-Savart law gives its size and direction; the total field is the integral over the wire.
- The field around a wire is made of circles centred on the wire. It has no component toward, away from or along the wire.
- A piece of wire aimed straight at the point (dℓ parallel to r̂) contributes nothing.
- When contributions point in different directions, use components and symmetry before integrating.
- The course expects calculations on the perpendicular bisector of a straight wire, on the axis of a circular loop, and at the centre of a loop or arc.
- μ₀ = 4π × 10⁻⁷ T·m/A, so μ₀/(2π) = 2 × 10⁻⁷ T·m/A.
Key relationships
| Situation | Result |
|---|---|
| Biot-Savart law | dB = (μ₀/4π) I dℓ sin θ / r², direction of dℓ × r̂ |
| Finite wire, length L, on its perpendicular bisector at distance a | B = μ₀IL / [2πa√(L² + 4a²)] |
| Long straight wire, distance r | B = μ₀I/(2πr) |
| Centre of a circular loop, radius R (N turns: multiply by N) | B = μ₀I/(2R) |
| Centre of an arc of angle φ in radians | B = μ₀Iφ/(4πR) |
| Axis of a loop, distance z from its centre | B = μ₀IR² / [2(R² + z²)^(3/2)]; ≈ μ₀IR²/(2z³) for z ≫ R |
| Force on a straight wire | F = Iℓ × B, size IℓB sin θ |
| Force on a curved wire | F = ∫ I dℓ × B |
| Two parallel wires, distance d | F/ℓ = μ₀I₁I₂/(2πd); same direction attract |
Assumptions behind the results
- Steady (constant) currents in thin wires.
- “Long” means the wire’s length is much greater than the distance to the point.
- The force results assume the field B is due to other sources; a wire exerts no net force on itself.
Mistakes to avoid
- B pointing away from the wire. It circles the wire.
- Forgetting sin θ. Leads aimed at the centre give zero field there.
- μ₀I/(2πR) at a loop’s centre. The loop result is μ₀I/(2R).
- Arc angle in degrees. Use radians.
- Adding magnitudes on the loop axis. Only the axial components survive.
- “Parallel currents repel.” They attract.
Quick self-check
- A quarter-circle arc of radius 0.20 m carries 5.0 A. What is B at its centre? (μ₀I/(8R) = 3.9 × 10⁻⁶ T)
- By what factor does B on the axis of a loop fall from z = 0 to z = R? (1/(2√2) ≈ 0.35)
- What is the net force on a closed loop in a uniform field? (Zero)
Next: practice questions.