3Magnetostatics

IB Electromagnetism

3.1 Ampere’s Law

Consider a surface

S

with boundary

C

. Current

J

flows through

S

. We now

integrate the first equation over the surface S to obtain

Z

S

(∇ × B) · dS =

I

C

B · dr = µ

0

Z

S

J · dS.

So

Law (Ampere’s law).

I

C

B · dr = µ

0

I,

where I is the current through the surface.

Example

(A long straight wire)

.

A wire is a cylinder with current

I

flowing

through it.

We use cylindrical polar coordinates (

r, ϕ, z

), where

z

is along the direction

of the current, and r points in the radial direction.

I

S

z

r

By symmetry, the magnetic field can only depend on the radius, and must lie

in the

x, y

plane. Since we require that

∇ · B

= 0, we cannot have a radial

component. So the general form is

B(r) = B(r)

ˆ

ϕ.

To find

B

(

r

), we integrate over a disc that cuts through the wire horizontally.

We have

I

C

B · dr = B(r)

Z

2π

0

r dϕ = 2πrB(r)

By Ampere’s law, we have

2πrB(r) = µ

0

I.

So

B(r) =

µ

0

I

2πr

ˆ

ϕ.

Example

(Surface current)

.

Consider the plane

z

= 0 with surface current

density k (i.e. current per unit length).

Take the x-direction to be the direction of the current, and the

z

direction to be

the normal to the plane.

We imagine this situation by considering this to be infinitely many copies of

the above wire situation. Then the magnetic fields must point in the

y

-direction.

By symmetry, we must have

B = −B(z)

ˆ

y,

with B(z) = −B(−z).

Consider a vertical rectangular loop of length L through the surface

L

Then

I

C

B · dr = LB(z) − LB(−z) = µ

0

kL

So

B(z) =

µ

0

k

2

for z > 0.

Similar to the electrostatic case, the magnetic field is constant, and the part

parallel to the surface is discontinuous across the plane. This is a general result,

i.e. across any surface,

ˆ

n × B

+

−

ˆ

n × B

−

= µ

0

k.

Example

(Solenoid)

.

A solenoid is a cylindrical surface current, usually made

by wrapping a wire around a cylinder.

z

r

C

We use cylindrical polar coordinates with

z

in the direction of the extension of

the cylinder. By symmetry, B = B(r)

ˆ

z.

Away from the cylinder,

∇ × B

= 0. So

∂B

∂r

= 0, which means that

B

(

r

)

is constant outside. Since we know that

B

=

0

at infinity,

B

=

0

everywhere

outside the cylinder.

To compute

B

inside, use Ampere’s law with a curve

C

. Note that only the

vertical part (say of length

L

) inside the cylinder contributes to the integral.

Then

I

C

B · dr = BL = µ

o

INL.

where

N

is the number of wires per unit length and

I

is the current in each wire

(so INL is the total amount of current through the wires).

So

B = µ

0

IN.