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.

B = μ0I⌠
⌡		ds × r̂
4πr2

line of current


[magnify]

+∞ +∞
Bline =  µ0I ⌠
⌡		 ds × r̂  =  µ0I ⌠
⌡		 y/√(x2 + y2)  dx k̂ =  µ0I ⌠
⌡		 y  dx k̂
4π r2 4π x2 + y2 4π (x2 + y2)3/2
−∞ −∞
+∞
Bline =  µ0I ⎡
⎢
⎣		 x ⎤
⎥
⎦		  k̂ =  µ0I ⎡
⎢
⎣		 +1  −  −1 ⎤
⎥
⎦		  k̂ =  µ0I 2 k̂
4π y(x2 + y2)½ 4π y y 4π y
−∞
Bline =  µ0I k̂
2πy

loop of current (along axis)


[magnify]

2π
Bloop =  µ0I ⌠
⌡		 ds × r̂  =  µ0I ⌠
⌡		 a/√(x2 + a2)  a dφ î
4π r2 4π x2 + a2
0
2π  
Bloop =  µ0I   a2 ⌠
⌡		 dφ î =  µ0I   a2  [2π − 0] î
4π (x2 + a2)3/2 4π (x2 + a2)3/2
0  
Bloop =  µ0I   a2  î
2 (x2 + a2)3/2

Solenoid

[solenoid pic goes here]

+∞
Bsolenoid =  ⌠
⌡		 dBloop =  µ0I ⌠
⌡		 a2  n dx î
2 (x2 + a2)3/2
−∞
+∞
Bsolenoid =  µ0nI ⎡
⎢
⎣		 x ⎤
⎥
⎦		  î =  µ0nI  [(+1) − (−1)] î
2 √(x2 + a2) 2
−∞

Bsolenoid = µ0nI î

ampère's law

Everything's better with Ampère's law (almost everything).

André Marie Ampère (1775-1836) France

∮B · ds = μ0I

∇ × B = μ0J

with displacement current term

∮B · ds = μ0ε0  ∂ΦB  + μ0I
dt
 
∇ × B = μ0ε0  ∂ΦE  + μ0J
dt

Apply to the straight wire, infinite sheet, solenoid, toroid,… anything else?

A straight wire - look how simple it is

∮B · ds  =  µ0I  
 
B(2πr)  =  µ0I  
 
B  =  µ0I  
2πr  

A solenoid - also wonderfully simple

∮B · ds  =  µ0I  
 
Bℓ  =  µ0NI  
 
B  =  µ0nI  
 

Beyond the straight wire lies the infinite sheet

∮B · ds  =  µ0I  
 
B(2ℓ)  =  µ0σℓ
B  =  µ0σ  
2  

Beyond the solenoid lies the toroid

∮B · ds  =  µ0I  
 
B(2πr)  =  µ0NI  
B  =  µ0NI  
2πr  

Inside a wire with total current I.

∮B · ds  =  µ0I  
 
B(2πr)  =  µ0I πr2
πR2
B  =  µ0Ir  
2πR2  

Inside a wire with charge density ρ.

∮B · ds = µ0I = µ0 ⌠
⌡		  ρ dV

The Physics Hypertextbook

  • No condition is permanent.