A brief note on Biot-Savart and Ampere's Circuital Law . Read it before a test. Try each question at the end before you open its answer.
The Biot-Savart law A piece of current: d B = μ 0 4 π I d l sin θ r 2 dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2} d B = 4 π μ 0 r 2 I d l sin θ , with μ 0 4 π = 10 − 7 T m/A \dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m/A} 4 π μ 0 = 1 0 − 7 T m/A . The field is along d l ⃗ × r ⃗ d\vec{l}\times\vec{r} d l × r . Direction: right-hand rule. Thumb along the current, fingers along the field. Wire of any length: B = μ 0 I 4 π d ( sin φ 1 + sin φ 2 ) B = \dfrac{\mu_0 I}{4\pi d}(\sin\varphi_1 + \sin\varphi_2) B = 4 π d μ 0 I ( sin φ 1 + sin φ 2 ) . Very long wire: B = μ 0 I 2 π d B = \dfrac{\mu_0 I}{2\pi d} B = 2 π d μ 0 I . Arc, at its centre: B = μ 0 I θ 4 π R B = \dfrac{\mu_0 I\theta}{4\pi R} B = 4 π R μ 0 I θ . Full loop of N N N turns: B = μ 0 N I 2 R B = \dfrac{\mu_0 N I}{2R} B = 2 R μ 0 N I . Axis of a loop: B = μ 0 N I R 2 2 ( R 2 + x 2 ) 3 / 2 B = \dfrac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} B = 2 ( R 2 + x 2 ) 3/2 μ 0 N I R 2 . Ampere's circuital law The law: ∮ B ⃗ ⋅ d l ⃗ = μ 0 I e n c \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc} ∮ B ⋅ d l = μ 0 I e n c . Only the net current through the loop counts. Long wire: B = μ 0 I 2 π r B = \dfrac{\mu_0 I}{2\pi r} B = 2 π r μ 0 I . Inside a round wire with an even current: B = μ 0 I r 2 π a 2 B = \dfrac{\mu_0 I\,r}{2\pi a^2} B = 2 π a 2 μ 0 I r . It is zero on the axis and largest at the surface. Inside a hollow pipe: B = 0 B = 0 B = 0 . Zero circulation does not mean zero field. Solenoid, toroid and parallel wires Long solenoid: B = μ 0 n I B = \mu_0 n I B = μ 0 n I inside, with n = N / L n = N/L n = N / L . At an end the field is half as much. Outside it is almost zero. Toroid: B = μ 0 N I 2 π r B = \dfrac{\mu_0 N I}{2\pi r} B = 2 π r μ 0 N I inside the ring. In the hole and outside it, B = 0 B = 0 B = 0 . Parallel wires: F L = μ 0 I 1 I 2 2 π d \dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} L F = 2 π d μ 0 I 1 I 2 on each metre. Same direction attracts. Opposite direction repels. A worked example A solenoid has 1000 1000 1000 turns on a tube 25 25 25 cm long and carries 0.60 0.60 0.60 A. Find the field inside.
Turns in each metre. n = 1000 0.25 m = 4000 m − 1 n = \dfrac{1000}{0.25\ \text{m}} = 4000\ \text{m}^{-1} n = 0.25 m 1000 = 4000 m − 1 . The field. B = μ 0 n I = 4 π × 10 − 7 T m/A × 4000 m − 1 × 0.60 A = 3.0 × 10 − 3 T B = \mu_0 n I = 4\pi \times 10^{-7}\ \text{T m/A} \times 4000\ \text{m}^{-1} \times 0.60\ \text{A} = 3.0 \times 10^{-3}\ \text{T} B = μ 0 n I = 4 π × 1 0 − 7 T m/A × 4000 m − 1 × 0.60 A = 3.0 × 1 0 − 3 T , that is 3.0 3.0 3.0 mT. Practise with new numbers
Check yourself The distance from a long straight wire is made three times larger, with the same current. What happens to the field? Stuck? Show a hint Show the answerTwo long parallel wires carry currents in the same direction. Do they attract or repel? Stuck? Show a hint Show the answerA toroid carries a current. What is the field in its hole? Stuck? Show a hint Show the answer