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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 × 
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 × 
r2
+∞
Bline =  µ0I
y/√(x2 + y2)  dx 
x2 + y2
−∞
+∞
Bline =  µ0I
y  dx 
(x2 + y2)3/2
−∞
+∞
Bline =  µ0I
x
 
y(x2 + y2)½
−∞
Bline =  µ0I
+1  −  −1
 
y y
Bline =  µ0I 2
y
Bline =  µ0I
y
B = μ0I
r

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 × 
r2
Bloop =  µ0I
a/√(x2 + a2)  a dφ 
x2 + a2
0
Bloop =  µ0I   a2
dφ 
(x2 + a2)3/2
0
Bloop =  µ0I   a2  [2π − 0] 
(x2 + a2)3/2
Bloop =  µ0I   a2  
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 
2 (x2 + a2)3/2
−∞
+∞
Bsolenoid =  µ0nI
x
2 √(x2 + a2)
−∞
Bsolenoid =  µ0nI  [(+1) − (1)] 
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 dΦE + μ0I
dt
∇ × 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
r

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
r

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
R2

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