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# 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. 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. 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

B = μ0nI

### ampère'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