Chapter 28 - Sources of Magnetic Fields
I.Introduction - we've already looked at how moving charges are affected by magnetic fields. Now let's take a look at the magnetic fields produced by the moving charges (current)
Two methods:1) Law of Biot-Savart
2) Ampere's Law
II.Magnetic Field Produced by a Moving Charge
We know that a point charge, q, produces an electric field. Does it also produce a magnetic field? The answer is yes!
A.Observations at point P a distance r from the charge that is traveling at a velocity v:
1.the magnetic field is proportional to the charge q, the velocity of the charge v, the reciprocal of the distance squared 1/r2, and the sine of the angle .
2.the direction of the magnetic field is perpendicular to the plane containing the velocity vector , and the distance vector -- in the direction of the cross product .
B.Expression for the magnetic field:
,
or,
where o = 4 x 10-7. Note that the units of the magnetic field is Tesla.
C.Interesting note: Remember = 9 x 109. Combine with o :
II.Law of Biot-Savart
A.Given a wire carrying a current I. What is the magnitude and direction of the magnetic field produced by the wire at a point P in space?
Start with a small segment of the wire and think of it as having many charges, q, flowing in it, with each charge contributing to the magnetic field at point P.
Consider a small segment of the wire of length dl that contains a total amount of charge dq. If n is the number of charges per unit volume, A is the cross-sectional area of the wire, and vd represents the drift velocity of the charges, then
dq = nAqdl.
The magnetic field produced by the small segment is like the magnetic field produced by a point charge. (Why?)
,
ordB =
Using vector notation:
which is the Law of Biot-Savart.
Note the magnitude and the direction of magnetic field produced by the small segment.
The magnetic field produced by a finite length of wire can be found by integrating over the entire length of the wire. Remember that the integration must be performed separately for the components of the magnetic field vectors.
B.Examples
1.Find the magnetic field a perpendicular distance a away from a “long” straight wire carrying a current I. Also, draw the lines of force of the magnetic field.
2.Two long parallel straight wires are carrying the same current I in the same direction. Find the magnetic field a distance y along the perpendicular bisector of the line connecting the two wires.
3.In question 2, find the magnitude and direction of the force exerted on the left wire due to the magnetic field produced by the right wire.
4.A long, flat metal ribbon of width w carries a current I. What is the magnitude and direction of the magnetic field at a distance y above the center of the ribbon?
5.Find the magnitude and direction of the magnetic field on the axis of a circular loop of wire of radius R and carrying a current I.
6.A flat disk of radius R has a charge Q uniformly distributed over its surface. The disk rotates about a vertical axis with an angular velocity . What is the magnitude and direction of the magnetic field at the center of the disk?
7.Find the magnitude and direction of the magnetic field along the axis of a solenoid - wire wound very close together in a helix. The solenoid has n turns per unit length of the solenoid. The radius of the helix is R and the current in the wires is I.
If the solenoid is closely wound and is very long (the length > radius), the solenoid is an ideal solenoid. In this case, the magnetic field is constant across the cross-sectional of the solenoid.
a.Find the magnetic field inside a long solenoid.
b.Find the magnetic field at one end of a long solenoid.
III.Ampere's Law - another way of finding the magnetic field for simple wire arrangements.
A.Come up with an expression for Ampere's law. Consider a long wire carrying a current I. Take a closed path around the wire and calculate along the path.
B.Examples
1.A long wire of radius R carries a current I that is uniformly distributed throughout the cross-section of the wire. Use Ampere's law to find the magnetic field both outside and inside the wire.
a.outside, rR
b.inside, rR
2.A coaxial cable with an inner conductor of radius a and an outside conductor of inside radius b and outside radius c. The inner conductor carries a current I in one direction and the outer conductor carries a current I in the opposite direction. Find the magnetic field between the two conductors (arb) and outside the outer conductor (rc).
3.Use Ampere's law to find the magnetic field inside an ideal (B field is uniform across the cross-section of the solenoid) solenoid with n turns of wire per unit length of the solenoid. The wires carry a current I.
4.Qualitative problem - introduce Maxwell's displacement current Id .
Apply Ampere's law to a path around the left wire.
a.The current passes through the areas A1and A2whose boundary is the path.
b.Is there a resolution to the problem?
IV.Magnetism in Matter - the magnetic effects in matter are due to magnetic moments. We already know that a current loop has a magnetic moment, and that a current loop is simply charge that is traveling in a circular path. So anytime we have charge that is traveling in a circular path, we have a magnetic moment. There are three charge motions that contribute to the magnetic moment of an atom.
A.Orbital motion of the electron. The motion of the electron around the nucleus of an atom constitutes a current loop and, therefore, has a magnetic moment, . It also has an angular momentum, L.
B.Spin of the electron - the electron can be thought of as spinning around an axis (a classical description) and this motion constitutes charge traveling around a circular path. Therefore, there is a magnetic moment associated with the spin.
C.Spin of the nucleons - again the nucleons can be thought of as a charged objects spinning, and they also have magnetic moments
D.The primary magnetic effects of atoms is a result of the spin magnetic moment of the electron. The orbital magnetic moments generally cancel, and the spin magnetic moments may or may not pair off and cancel. Generally, those atoms with an even number of electrons tend to have no magnetic moment while those atoms with an odd number of electrons have a net magnetic moment.
V.Magnetic Properties
A.Examine two solenoids, one without any material inside and one with a material inside.
without materialwith material
The total magnetic field inside a solenoid (without or with a material) is written as , where .
B. is the magnetic field due to the current in the wires of the solenoid. It can be written as , where = the magnetic field strength in A/m and is dependent on the current in the wires (e.g., remember for a solenoid B = onI = oH, hence H = nI).
C. is the magnetic field contribution due to the material and can be written as is called the magnetization of the material and equals sum of the magnetic moments per unit volume .
D.Therefore, in general,
.
E.The magnetization should also be dependent on the magnetic field produced by the wires of the solenoid because the magnetic field will tend to align the magnetic moments (remember ). Therefore, is dependent on the magnetic intensity and we can then write The total magnetic field in the solenoid can be rewritten as
.
The constants are = magnetic susceptibility and = permeability of the material. A third constant called the relative permeability is defined as
This is the magnetic equivalent of the electrical dielectric constant. Some values of the magnetic susceptibility and the relative permeability are given in the table below:
material / / m =vacuum / 0 / 1
aluminum / 2.3 x 10-5 / 1.000023
copper / -9.8 x 10-6 / 0.9999902
iron / 10 to 100,000 / 10 to 100,000
Note how large (or small) the field with the material is in comparison to the field without the material.
VI.Types of Magnetic Materials - paramagnetic, diamagnetic, ferromagnetic materials
A.Paramagnetic materials exhibit paramagnetism. These are materials where m > 1, or the total magnetic field is greater than the magnetic field due to just the solenoid current (BBo). This means that the material produces a small magnetic field that is in the same direction as the applied field.
1.Examples: aluminum, lithium, platinum, etc. All these atoms have a net magnetic moment.
2.Apply an external magnetic field to a paramagnetic material:
a.The magnetic torque tends to align magnetic moments with the external field. This alignment causes an increase in the total magnetic field.
b.We expect to get all the magnetic moments aligned and as a result have a large increase in the magnetic field. However, thermal effects prevent this from occurring. Curie's Law describes the magnetization as a function of the absolute temperature.
B.Ferromagnetic materials exhibit ferromagnetism. These are materials where m > 1, which means that these materials produce a very large increase in the total magnetic field (BBo ).
1.Examples: iron, cobalt, nickel, etc. The atoms in these materials have a magnetic moment, but the strong effect is due to many of these atoms forming "domains" where their magnetic moments all point in the same direction (by "exchange coupling"). The size of these "domains" is about 10-5 cm.
2.Ferromagnetic materials that are not yet magnetized have domains where their magnetic moments point in random directions.
3.When placed in an external magnetic field, the magnetic moments of the domains tend to line up with the external magnetic field by domain growth and domain rotation.
Mechanism of domain growth: What is happening at the boundary?
4.Magnetization M vs Magnetic Intensity H (which is proportional to the current I )
5.Hysteresis: Place a hunk of ferromagnetic material in a solenoid. Gradually increase the current in the solenoid to some value, then gradually decrease the current back to zero. Then gradually increase the current again, but in the opposite direction (this reverses the B field in the solenoid). Then decrease it back to zero. This process can be repeated over and over again.
The changing direction of the magnetic field changes the direction of the magnetic moments in the material and thus, requires energy. This energy is converted into heat energy and is directly related to the area of the “hysteresis” curve.
Problem: Draw a hyteresis curve for a material that 1) could be a good candidate for a permanent magnet, 2) might make a strong permanent magnet but can be easily demagnetized, 3) can produce a magnetic field (even though it may be weak) but cannot be easily demagnetized, and 4) minimizes the losses of energy in a transformer.
C..Diamagnetic materials exhibit diamagnetism. These are materials where m < 1, or the total magnetic field is less than the field produced by the solenoid current (BBo ). This means that these materials produce a small magnetic field that is in the opposite direction to the applied field.
1.Examples: copper, gold, mercury, etc. These atoms do not have a net magnetic moment.
2.When placed in an external magnetic field, a weak magnetic moment is set up in the opposite direction to the applied field, thus resulting in a total field that is less than the field due just to the current. The effect, however, is very small and is masked by paramagnetic and ferromagnetic effects (B = Bo - Bm ). The oppositely directed magnetic field is an induced magnetic field (Faraday’s Law).
VIII. Gauss's Law for Magnetism - Reprise
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