Electromagnetism - Lecture 3
Magnetic Fields
• Magnetic Fields
• Integral form of Ampere’s Law
• Differential form of Ampere’s Law
• Magnetic Vector Potential
• Methods of calculating Magnetic Fields
• Examples of Magnetic Fields
1
Magnetic Field
The magnetic field B is defined by the force on a moving charge:
in units of Tesla, T=NA−1 m−1
F = qv × B
Force on a current element:
dF = Idl × B = J × Bdτ
The directions of F, B and dl using the left-hand rule:
B is in the direction of the thumB
Idl is in the direction of the Index finger
F is in the direction of motion and of the Middle finger
2
Ampère’s Law for B
The integral of the magnetic field round a closed loop is related to
the total current flowing across the surface enclosed by the loop.
I
B.dl = µ0 I = µ0
L
Z
J.dS
A
Integral must be round a closed loop!
Lines of magnetic flux are closed loops around currents.
Direction of B relative to I given by corkscrew rule
For symmetric problems can assume that |B| is the same at all
points around loop
3
Differential Form of Ampère’s Law
Applying Stokes’s theorem to the integral of the magnetic field
round a closed loop:
Z
Z
I
B.dl =
∇ × B.dS = µ0
J.dS
L
A
A
Removing the integral over the area gives the differential form of
Ampère’s Law:
∇ × B = µ0 J
At any point in space the curl of the magnetic field is proportional
to the local current density
In electrostatics the equivalent statement for the electric field is:
∇ × E = −∇ × ∇V = 0
4
Magnetic Vector Potential
Because ∇.B = 0, the magnetic field can always be expressed as
the curl of a magnetic vector potential A (“div curl =0”):
B=∇×A
∇.B = ∇.(∇ × A) = 0
Using Stokes’s theorem over a closed loop gives an integral
relationship:
Z
I
A.dl =
B.dS = ΦB
L
A
The magnetic flux through a surface is the integral of the magnetic
vector potential around the loop enclosing the surface.
Direction of a contribution dA is parallel to a current element Idl
5
Poisson’s Equation for A
Replacing B with ∇ × A in the differential form of Ampère’s Law:
∇ × B = ∇ × ∇ × A = µ0 J
Using the identity “curl curl = grad div - delsquared”:
∇2 A − ∇(∇.A) = −µ0 J
The choice of ∇.A = 0 is made for static fields
This is known as the Coulomb gauge
It leads to Poisson’s equation for A:
∇2 A = −µ0 J
At any point in space the second derivatives of the magnetic vector
potential are proportional to the local current density
6
Calculating Magnetic Fields
1. By summing the contributions to B as components of vectors
µ0 (J × r̂)dτ
µ0 I(dl × r̂)
=
dB =
4πr 2
4πr 2
This is known as Biot-Savart’s Law
2. By summing the contributions to A as components of vectors
µ0 Jdτ
µ0 Idl
=
dA =
4πr
4πr
and then taking the curl B = ∇ × A
Note that there is no method summing a scalar potential
3. By using Ampère’s Law
Z
I
B.dl = µ0 I = µ0
J.dS
L
A
For symmetric problems and uniform charge distributions
7
Notes:
Diagrams:
8
Magnetic Field of an Infinite Wire
Apply Ampère’s Law to a circular loop around wire
By symmetry there is only a component Bφ around the wire
Outside a wire carrying a current I there is a 1/r field:
µ0 I
2πr
and the magnetic vector potential is obtained from the only
non-zero contribution to the curl:
∂Az
µ0 I
Az = −
ln(r)
Bφ = −
∂r
2π
Bφ 2πr = µ0 I
Bφ =
Inside a wire with uniform current density J:
µ0 Ir 2
µ0 Jr 2
Az = −
=−
2
4πR
4
µ0 Ir
µ0 Jr
=
Bφ =
2πR2
2
9
Magnetic Field of a Current Loop
Along the axis of the loop only a Bz component by symmetry
Using Biot-Savart’s Law the contributions to Bz are:
µ0 Idl sin θ
dBz =
4πr 2
where the sin θ comes from dl × r̂
Integrating round the loop:
µ0 I2πa sin θ
µ0 Ia2
=
Bz =
4πr 2
2(a2 + z 2 )3/2
At the centre of the loop:
Bz (z = 0) = µ0 I/2a
At a large distance from the loop there is a 1/z 3 field:
µ0 Ia2
Bz (z >> a) =
2z 3
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Notes:
Diagrams:
11
Notes:
Diagrams:
12
Magnetic Field of a Solenoid
A solenoid consists of n circular current loops per unit length.
Calculate axial Bz from sum of individual loops:
Z
µ0 nIa2
dz
Bz =
2
2
3/2
4(a + z )
Trick is to change from dz to integral over θ:
Z
µ0 nI
sin θdθ
Bz =
2
For an infinite solenoid there is a uniform axial field:
Bz = µ0 nI
This result can also be shown using Ampere’s Law
13
Magnetic Field of a Toroid
A toroid is a solenoid bent in a large circle of radius R
It looks like a doughnut with a hole in it
There are n circular loops per unit length round the large circle
Use Ampere’s Law round a loop of radius R:
Bφ 2πR = µ0 (n2πR)I
Bφ = µ0 nI
Inside and outside the circular loops B = 0 from Ampere’s Law
The magnetic field is contained inside the loops of the toroid
14
Notes:
Diagrams:
15
Notes:
Diagrams:
16