Sources of Magnetic Field

Magnetic Field of a Moving Charge Recall the electric field E due to a charged particle (like a test charge) at point P at a distance r from the charge :

Similar to an electric field, the magnetic field B is also proportional to |q| and 1/r2. However, the direction of B is not along the line from source point to field point. Instead, B is perpendicular to the plane containing this line and the particle’s velocity v. Thus the magnetic field magnitude at point P is given by:

!E = 1

4!"0qr2r̂

B = µ04!

q vsin"r2

!B = µ0

4!q !v " r̂r2

The magnetic field B in vector form is:

µ0 is the permeability of free space.µ0 = 4! "10#7N $ s2 C 2 = 4! "10#7Wb A $m = 4! "10#7T $m A

Biot-Savart Law

The principle of the superposition for magnetic field states that:

The total magnetic field caused by several moving charges is the vector sum of the fields caused by the individual charges. Hence,

dQ = nqAdl

dB = µ04!

dQ vd sin"r2

= µ04!

v q vdAdl sin"r2

I = n q vdA

dB = µ04!

Idl sin"r2

d!B = µ0

4!Id!l " r̂r2 Biot-Savart Law

!B = µ0

4!Id!l " r̂r2#

Magnetic Field of a Current Element

Question (1)

Two protons move parallel to the x-axis in opposite directions at the same speed v (small compare to c) as shown in the figure. At the instant shown, find

a)! find the electric force on the upper proton b)! find the magnetic force on the upper proton c)! determine the ratio of the magnitudes of the

two forces.

P

1

2

3

4

Question (2)

1) direction 1

2) direction 2

3) direction 3

4) direction 4

5) the B field is zero

If the currents in these wires have the same magnitude but opposite directions, what is the direction of the magnetic field at point P?

Ampère’s Law The magnetic field in space around an electric current is proportional to the electric current which serves as its source, just as the electric field in space is proportional to the charge which serves as its source. 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. current enclosed in the loop.

B!! "l = µ0IMore general form of Ampère’s law is given by:

!B !d!l"" = µ0Iencl .

Ampere’s law in magnetism is analogous to Gauss’s law in electrostatics. In order to apply them, the system must possess certain symmetry.

Ampere’s law is applicable to the following current configurations: 1.! Infinitely long straight wires carrying a steady current. 2.! Infinitely large sheet of thickness b with a current density. 3.! Infinite solenoid. 4.! Toroid.

Magnetic Field of a Current-Carrying Conductor: Ampère’s Law

Let’s calculate magnetic field using Ampère’s law :

Choose loop to be a circle of radius R centered on the wire in a plane ! to wire. Magnitude of B is constant and direction of B is parallel to dl.!

B= µ0

2!Ir

!B !d!l"" = µ0Iencl.

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d

d

B dl!! = B 2"r( ) = µ0Iencl . # B = µ0Iencl .

2"r

Question (3)

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1) +z (out of page)

2) -z (into page)

3) +x

4) -x

5) -y

A positive charge moves parallel to a wire. If a current is suddenly turned on, in which direction will the force act?

Question (4)

Consider a long straight wire of radius R carrying a current I of uniform current density, as shown in the figure. Find the magnetic field everywhere. (Field Inside & Outside of a Wire)

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Magnetic Field due to a Circular Current Loop

Calculating the magnetic field at an axial point P a distance x from the center of the loop.

d!B = µ0I

4!d!l " r̂r2

= µ0I4!

dla2 + x2( )

Note : the y components will cancel due to symmetry.

dBx =µ0I4!

dla2 + x2( ) cos"

cos" = aa2 + x2

#

$%%

&%%

' Bx = dBx!( = µ0I4!

dla2 + x2( )!(

a

a2 + x2( )12= µ0I4!

a

a2 + x2( )32dl!(

Bx =µ0I4!

a

a2 + x2( )322!a( ) = µ0Ia

2

2 a2 + x2( )32

Solenoid A solenoid is a long coil of wire tightly wound in the helical form. The figure shows the magnetic field lines of a solenoid carrying a steady current I.

For an “ideal” solenoid, which is infinitely long with turns tightly packed, the magnetic field inside the solenoid is uniform and parallel to the axis, and vanishes outside the solenoid.

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We can use Ampere’s law to calculate the magnetic field strength inside an ideal solenoid.

The cross-sectional view of an ideal solenoid is shown below. To compute B , we consider a rectangular path of length l and width w and traverse the path in a counterclockwise manner. The line integral of B along this loop is

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( CBEH

!B !d!l =

!B !d!l

1""" +

!B !d!l

2" +

!B !d!l

3" +

!B !d!l

4"

The contributions along sides 2 and 4 are zero because B is perpendicular to dl . In addition, B = 0 along side 1 because the magnetic field is non-zero only inside the solenoid. The total current enclosed by the Amperian loop is Ienc= NI , where N is the total number of turns. Applying Ampere’s law yields

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( CBEH

where n = N / l represents the number of turns per unit length.

!B"! "d!l = Bl = µ0NI or B = µ0NI

l= µ0nI

Solenoid

!B !d!l =

!B !d!l

1""" +

!B !d!l

2" +

!B !d!l

3" +

!B !d!l

4"

= 0 + 0 + Bl + 0

Toroid Consider a toroid which consists of N turns, as shown in the figure. Find the magnetic field everywhere.

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One can think of a toroid as a solenoid wrapped around with its ends connected. Thus, the magnetic field is completely confined inside the toroid and the field points in the azimuthal direction (clockwise due to the way the current flows, as shown in the figure). Applying Ampere’s law, we obtain

where r is the distance measured from the center of the toroid. Unlike the magnetic field of a solenoid, the magnetic field inside the toroid is non-uniform and decreases as 1/ r .

!B !d!s = Bds = B ds"""""" = B 2!r( ) = µ0NI

B = µ0NI2!r

Force Between Parallel Wires

Calculate the force on length L of wire b due to the field of wire a .

The magnetic field at b due to a is given by:

Ba =

µ0ia2!d

" !Fb = ia

!L #!Ba =

µ0iaibL2!d

Calculate the force on length L of wire a due to the field of wire b .

The magnetic field at a due to b is given by:

Bb =

µ0ib2!d

" !Fa = ib

!L #!Bb =

µ0ibiaL2!d

Magnetic Materials

!! Ferromagnetic materials are those that can become strongly magnetized, such as iron and nickel.

!! These materials are made up of tiny regions called domains; the magnetic field in each domain is in a single direction.

!! When the material is unmagnetized, the domains are randomly oriented. They can be partially or fully aligned by placing the material in an external magnetic field.

!! A magnet, if undisturbed, will tend to retain its magnetism. It can be demagnetized by shock or heat.

!! The relationship between the external magnetic field and the internal field in a ferromagnet is not simple, as the magnetization can vary.

Magnetic Materials: Ferromagnetism

fully aligned by placing the material in an external

is not