Ampère's Law
Discussion
biot-savart law
This law usually no fun to deal with, but it's the elementary basis (the most primitive statement) of electromagnetism. Jean-Baptiste Biot and Félix Savart.
| B = | μ0I | ⌠ ⎮ ⌡ | ds × r̂ |
| 4π | r2 |
Let's apply it to three relatively easy situations: a straight wire, a single loop of wire, and a coil of wire with many loops (a solenoid).
the straight wire
Given an infinitely long, straight, current carrying wire, use the Biot-Savart law to determine the magnetic field strength at any distance r away.
Start with the Biot-Savart Law because the problem says to.
| B = | μ0I | ⌠ ⎮ ⌡ | ds × r̂ |
| 4π | r2 |
| +∞ | ||||
| Bline = | μ0I | ⌠ ⎮ ⌡ |
y/√(x2 + y2) | dx k̂ |
| 4π | x2 + y2 | |||
| −∞ | ||||
| +∞ | ||||
| Bline = | μ0I | ⌠ ⎮ ⌡ |
y | dx k̂ |
| 4π | (x2 + y2)3/2 | |||
| −∞ | ||||
| +∞ | |||||
| Bline = | μ0I | ⎡ ⎢ ⎣ |
x | ⎤ ⎥ ⎦ |
k̂ |
| 4π | y(x2 + y2)½ | ||||
| −∞ | |||||
| Bline = | μ0I | ⎡ ⎢ ⎣ |
+1 | − | −1 | ⎤ ⎥ ⎦ |
k̂ |
| 4π | y | y |
| Bline = | μ0I | 2 | k̂ | |
| 4π | y |
| Bline = | μ0I | k̂ |
| 2πy |
| B = | μ0I |
| 2πr |
the single loop of wire
Given a current carrying loop of wire with radius a, determine the magnetic field strength anywhere along its axis of rotation at any distance x away from its center.
Start with the Biot-Savart Law because the problem says to.
| B = | μ0I | ⌠ ⎮ ⌡ | ds × r̂ |
| 4π | r2 |
| 2π | ||||
| Bloop = | μ0I | ⌠ ⎮ ⌡ |
a/√(x2 + a2) | a dφ î |
| 4π | x2 + a2 | |||
| 0 | ||||
| 2π | |||||
| Bloop = | μ0I | a2 | ⌠ ⎮ ⌡ |
dφ î | |
| 4π | (x2 + a2)3/2 | ||||
| 0 | |||||
| Bloop = | μ0I | a2 | [2π − 0] î | |
| 4π | (x2 + a2)3/2 |
| Bloop = | μ0I | a2 | î | |
| 2 | (x2 + a2)3/2 |
| B = | μ0I | a2 | |
| 2 | (x2 + a2)3/2 |
the solenoid
Given a coil with an infinite number of loops (an infinite solenoid), determine the magnetic field strength inside if the coil has n turns per unit length.
[solenoid pic goes here]
| Bsolenoid = | ⌠ ⌡ |
dBloop |
Strictly speaking, this isn't an application of the Biot-Savart law. It's really just an application of pure calculus. What is a solenoid but a stack of coils and an infinite solenoid is an infinite stack of coils. Calculus loves infinity. It eats it for breakfast.
| +∞ | ||||
| Bsolenoid = | μ0I | ⌠ ⎮ ⌡ |
a2 | n dx î |
| 2 | (x2 + a2)3/2 | |||
| −∞ | ||||
| +∞ | ||||
| Bsolenoid = | μ0nI | ⎡ ⎢ ⎣ |
x | ⎤ ⎥ ⎦ |
| 2 | √(x2 + a2) | |||
| −∞ | ||||
| Bsolenoid = | μ0nI | [(+1) − (−1)] î |
| 2 |
Bsolenoid = μ0nI î
B = μ0nIampère's law
Everything's better with Ampère's law (almost everything).
André-Marie Ampère (1775–1836) France
The law in integral form.
∮B ⋅ ds = μ0I
The law in differential form.
∇ × B = μ0J
These forms of the law are incomplete. The full law has an added term called the displacement current. We'll discuss what all of this means in a later section of this book. For now, just look at the pretty symbols.
| ∮B ⋅ ds = μ0ε0 | ∂ΦE | + μ0I |
| ∂t |
| ∇ × B = μ0ε0 | ∂E | + μ0 J |
| ∂t |
Apply to the straight wire, flat sheet, solenoid, toroid, and the inside of a wire.
the straight wire
A straight wire. Look how simple it is.
[straight wire with amperean path goes here]
Start with Ampère's law because it's the easiest way to derive a solution.
∮B ⋅ ds = μ0I
B(2πr) = μ0I
| B = | μ0I |
| 2πr |
the flat sheet
Beyond the straight wire lies the infinite sheet.
[infinite sheet with amperean path goes here]
Start with Ampère's law because it's the easiest way to derive a solution.
∮B ⋅ ds = μ0I
B(2ℓ) = μ0σℓ
| B = | μ0σ |
| 2 |
the solenoid
A solenoid. Also wonderfully simple.
[solenoid with amperean path goes here]
Start with Ampère's law because it's the easiest way to derive a solution.
∮B ⋅ ds = μ0I
Bℓ = μ0NI
B = μ0nI
the toroid
Beyond the solenoid lies the toroid.
[toroid with amperean path goes here]
Watch me pull a rabbit outta my hat, starting with Ampère's law because it's the easiest way to pull a rabbit out of a hat.
∮B ⋅ ds = μ0I
B(2πr) = μ0NI
| B = | μ0NI |
| 2πr |
the inside of a wire
What's it like to be inside a wire — inside a wire with total current I?
[amperean path inside a wire goes here]
Start with Ampère's law because it's the easiest way to arrive at a solution.
∮B ⋅ ds = μ0I
| B(2πr) = μ0I | πr2 |
| πR2 |
| B = | μ0Ir |
| 2πR2 |
What's it like to be inside a wire — inside a wire with current density ρ?
Back to Ampère's law one last time.
∮B ⋅ ds = μ0I
B(2πr) = μ0ρ(πr2)
| B = | μ0ρr |
| 2 |