Magnetostatics
Introduit| English | Sens |
|---|---|
| Biot–Savart law/ˈbɪɒt ˈsævɑːt lɔː/ | Biot–Savart law |
| Ampère's law/ˈæmpeəz lɔː/ | The magnetic field around a closed loop is set by the total current passing through the loop. |
| inductance/ɪnˈdʌktəns/ | inductance |
| magnetic flux/mæɡˈnetɪk flʌks/ | A measure of the magnetic field passing through a surface, accounting for field strength, area and orientation. |
| Lorentz force | force de Lorentz |
Where motors begin
- Motors, relays, MRI scanners and every inductor start from one fact: currents make magnetic fields.
- For a general shape you must add up every bit of current. When the shape is symmetric, one line of algebra does it. This lesson does both.
Two laws for one field
- The Biot–Savart law 毕奥萨伐尔定律 adds the field from every small piece of current. It always works, but usually means a hard integral.
- Ampère's law 安培环路定律 wraps the field round the current: $\displaystyle\oint \mathbf{B} \cdot d\mathbf{l} = \mu_{0}I_{enc}$. With the right symmetry, the loop integral becomes $B \times$ length.
When is Ampère's law the quick route to B?
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
The law always holds, but it gives $B$ directly only when the loop integral reduces to $B \times$ length. Otherwise use Biot–Savart.
A long straight wire carries 10 A. What is B at 5 cm from it, in µT?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
A circle round the wire: $B \cdot 2\pi r = \mu_{0}I$, so $B = \dfrac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.05} = 40\,\mu$T.
The infinite solenoid
- Take a rectangular loop with one side of length $l$ inside the solenoid and one outside. It encloses $nlI$ of current, with $n$ turns per metre.
- Outside, the field cancels; the short sides cross the field at right angles. So $Bl = \mu_{0}nlI$, and $B = \mu_{0}nI$ inside, uniform.
Worked example: the field inside
- $n = 1000$ turns/m, $I = 2\,\text{A}$: $B = 4\pi \times 10^{-7} \times 1000 \times 2 = 2.51\,\text{mT}$, the same everywhere along the axis.
- Biot–Savart would mean integrating a thousand loops. Checking symmetry first saves a page of work.
A long solenoid has 1000 turns/m and carries 2 A. The field inside is about
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
$B = \mu_{0}nI = 4\pi \times 10^{-7} \times 1000 \times 2 = 2.51$ mT, uniform along the axis.
Flux and inductance
- Magnetic flux 磁通量 is $\Phi = \displaystyle\int \mathbf{B} \cdot d\mathbf{A}$, and inductance 电感 is flux linkage per amp: $L = \dfrac{N\Phi}{I}$.
- For a solenoid of cross-section $A$: $\dfrac{L}{l} = \mu_{0}n^{2}A$. Inductance grows with the square of the turns density.
Doubling the turns per metre of a solenoid doubles its inductance per metre.
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
$\dfrac{L}{l} = \mu_{0}n^{2}A$: doubling $n$ doubles the field and doubles the turns it links, so $L$ rises four times.
The Lorentz force
- A charge moving through a field feels the Lorentz force 洛伦兹力 $\mathbf{F} = q\mathbf{v} \times \mathbf{B}$, at right angles to both. It turns the charge without speeding it up.
- So the charge moves in a circle of radius $r = \dfrac{mv}{qB}$.
An electron at $10^{6}\,\dfrac{\text{m}}{\text{s}}$ in 1 T: $r = \dfrac{9.11 \times 10^{-31} \times 10^{6}}{1.6 \times 10^{-19} \times 1} = 5.7\,\mu\text{m}$. Its energy is 2.85 eV, far below relativistic, so the formula holds.
An electron at 10⁶ m/s moves at right angles to a 1 T field. Its orbit radius is about
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
$r = \dfrac{mv}{qB} = \dfrac{9.11 \times 10^{-31} \times 10^{6}}{1.6 \times 10^{-19}} = 5.7\,\mu$m.
A 1 µC charge moves at 10⁶ m/s at right angles to a 1 T field. What force does it feel, in newtons?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
$F = qvB = 10^{-6} \times 10^{6} \times 1 = 1$ N, at right angles to both the velocity and the field.
Three common slips
- Forgetting the turns. A loop cutting $N$ windings encloses $NI$, not $I$.
- Treating a short coil as infinite. $B = \mu_{0}nI$ needs length much greater than diameter.
- Expecting B to be continuous at a current sheet. Its tangential part jumps by $\mu_{0}K$.
Match each quantity to its formula.
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
Each comes from Ampère's law or the Lorentz force with one symmetry argument.
Recap
- Ampère's law with symmetry: $Bl = \mu_{0}I_{enc}$.
- Infinite solenoid: $B = \mu_{0}nI$ inside, $\dfrac{L}{l} = \mu_{0}n^{2}A$.
- Lorentz force $q\mathbf{v} \times \mathbf{B}$: circles of radius $\dfrac{mv}{qB}$.