# Ampere's Law Differential Form

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## Discussion

Physics - E&M: Maxwell's Equations (28 of 30) Faraday's Law in Differential Form Ex. 1 - Duration: 8:51. Michel van Biezen 14,095 views. What is a Transformer And How Do They Work? In this article, we shall study the Ampere’s law and its application in finding magnetic induction due to long straight conductor, solenoid and toroid. Statement: The line integral of the magnetic field B around any closed path is equal to μ 0 times the net steady current enclosed by this path. Ampere's Law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.

### 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) dxk̂ 4π x2 + y2 −∞
 +∞ Bline = μ0I ⌠⌡ y dxk̂ 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 dxî 2 (x2 + a2)3/2 −∞
 +∞ Bsolenoid = μ0nI ⎡⎢⎣ x ⎤⎥⎦ 2 √(x2 + a2) −∞

### Ampere's Circuital Law Differential Form

 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 + μ0J ∂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]

### Ampere's Law Differential Form

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

### Ampere's Law Examples

B(2πr) = μ0NI

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 B = μ0NI 2πr

### Ampere's Force Law

#### 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

### Ampere's Law

 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

### Ampere S Law Calculator B(2πr) = μ0ρ(πr2)

 B = μ0ρr 2