Chapter 28

Sources of Magnetic Field

In this chapter we investigate the sources of magnetic of magnetic field, in particu-lar, the magnetic field produced by moving charges (i.e., currents). Ampere’s Lawis introduced and plays a role analogous to Gauss’s Law. At the end of the chapter,various forms of magnetism produced in materials at the atomic and particle levelare discussed (i.e., paramagnetism, diamagnetism, and ferromagnetism).

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1 Magnetic Field of a Moving Charge

Let’s investigate the magnetic field due to a single point charge q moving with aconstant velocity ~v. As we did for electric fields, the location of the moving chargeat a given instant will be called the source point, and the point P where we wantto calculate the magnetic field will be called the field point.

Figure 1: This figure shows the magnetic-field vectors due to a moving positive point charge q. Ateach point, ~B is perpendicular to the plane of ~r and ~v. Notice that ~B points in the direction of~v × ~r.

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1.1 Moving Charge: Vector Magnetic Field

~B =µo4π

q ~v × rr2

(1)

where µo = 4π × 10−7 T·m/A (exactly).

A point charge in motion also produces an electric field, with field lines pointingradially outward from a positive charge. The magnetic field lines are completelydifferent. The magnetic field lines are circles centered on the line of ~v and lying inplanes perpendicular to this line.

When we study the propagation of electromagnetic waves we will see that theirspeed of propagation is related to the electric permittivity and the magnetic per-meability by the following relation:

c2 =1

εoµo

The Forces Between Two Moving Protons

Figure 2: This figure shows the electric and magnetic forces acting on a pair of protons moving inopposite directions at their point of closest approach.

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~FE =1

4πεo

q2

r2

~B =µo4π

q(v ı)× r2

=µo4π

qv

r2k

The velocity of the upper proton is −~v = −v i, so, the magnetic force on it is:

~FB = q(−~v)× ~B = q(−vı)× µo4π

qv

r2k =

µo4π

q2v2

r2

The magnetic interaction in this situation is also repulsive. The ratio of the forcemagnitudes is

FB

FE= εoµo v

2 =v2

c2

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2 Magnetic Field of a Current Element

Figure 3: This figure shows the magnetic-field vectors due to a current element d~. The view alongthe current axis is also shown.

dQ = nqAd`

The moving charges in this segment are equivalent to a single charge dQ travelingwith a velocity equal to the drift velocity ~vd.

dB =µo4π

|dQ| vd sinφ

r2=

µo4π

n|q| vdAd` sinφ

r2

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dB =µo4π

I d` sinφ

r2(2)

Current Element: Vector Magnetic Field

This can be written in a more compact vector notion:

d ~B =µo4π

I d~× rr2

(3)

The total magnetic field due to a current-carrying conductor is the sum of theindividual d ~B contributions:

~B =µo4π

∫I × d~× r

r2(Biot-Savart Law) (4)

3 Magnetic Field of a Straight Current-Carrying Conduc-

tor

Figure 4: This figure shows the magnetic field produced by a straight current-carrying conductorof length 2a.

Using the Bio-Savart Law, we can calculate the magnetic field produced by astraight current-carrying conductor.

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B =µoI

4π

∫ a

−a

x dy

(x2 + y2)3/2

Integrating this by trigonometric substitution (y = x tan θ), we obtain:

B =µoI

4π

2a

x√x2 + a2

(due to a finite wire of length 2a) (5)

If we extend this solution for an infinitely long wire by taking the limit that a→∞,we find the following magnetic field a distance r perpendicular the wire is:

B =µoI

2πr(due to an infinitely long wire)

The magnetic field B has an axial symmetry about the y-axis (see Fig. 5 below)

Magnetic Field of a Single Wire

B =µoI

2πr(6)

Figure 5: This figure shows the magnetic-field around a long, straight, current-carrying conductor.The field lines form concentric circles, with directions determined by the right-hand rule.

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Magnetic Field of Two Wires

What is the magnetic field resulting from two long, straight current-carrying wires?We just use the principel of superposition. That is, ~Btotal = ~B1 + ~B2.

Figure 6: This figure shows the magnetic-field lines for two long, straight conductors carrying equalcurrents in opposite directions. The magnetic field is strongest in the region between the conductors.

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4 Force Between Parallel Conducting Wires Carrying Cur-

rent

Figure 7: This figure shows how the magnetic field ~B caused by the current in the lower conductorexerts a force ~F on the upper conductor. The force between two parallel wires carrying current inthe same direction are attractive, and consistent with Newton’s 3rd law.

The magnetic field ~B produced by the lower conductor is:

B =µoI

2πr

The force this field exerts on a length L of the upper wire is ~F = I ′ ~L× ~B, wherethe vector ~L is in the direction of the current I ′.

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F = I ′LB =µoII

′ L

2πr

This can be written in terms of the force per-unit-length:

F

L=

µo I I′

2πr(7)

Magnetic Forces and Defining the Ampere

One ampere is that unvarying current that, if present in eachof two parallel conducting wires of infinite length and one me-ter apart in empty space, causes each conductor to experiencea force of exactly 2× 10−7 newtons per meter of length.

5 Magnetic Field of a Circular Current Loop

Figure 8: This figure shows the magnetic-field vector d ~B produced by a current element I d~ withcomponents resolved along the x and y axes.

We use Eq. 3 to determine the “magnitude of d ~B” produced by a current elementI d~ and find the following:

dB =µoI

4π

dl

(x2 + a2)

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Next, the components of the d ~B vector are resolved along the x and y directions,and we find the following:

dBx = dB cos θ =µoI

4π

dl

(x2 + a2)

a

(x2 + a2)1/2

dBy = dB sin θ =µoI

4π

dl

(x2 + a2)

x

(x2 + a2)1/2

Integrating dBy gives a null result. Meanwhile, integrating the x-component of dBgives:

Bx =

∫dBx =

∫µoI

4π

a d`

(x2 + a2)3/2=

µoI

4π

a

(x2 + a2)3/2

∫d`

The integration∫d~= 2πa gives the follow result:

Bx =µoIa

2

2 (x2 + a2)3/2(Magnetic field on the axis) (8)

Magnetic Field on the Axis of a Coil

Figure 9: This figure shows how to apply the right-hand rule for the direction of the magnetic fieldproduced on the axis of a current-carrying coil.

The above equation can be written in terms of the magnetic dipole moment µ,where µ = NIπa2 where N is the number of turns in the coil, and πa2 is the areaof the coil.

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Bx =µo µ

2π (x2 + a2)3/2

Figure 10: This figure shows the magnetic field strength along the axis of a circular coild with Nturns. When x is much larger than a, the field magnitude decreases approximately as 1/x3.

Figure 11: This figure shows the magnetic field lines produced by the current in a circular loop.The ~B field along the x-axis has the same direction as the magnetic moment of the current loop ~µ.

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6 Ampere’s Law

Gauss’s law for electric fields involves the flux of ~E through a closed surface, and itstates that the flux is equal to the total charge enclosed within the surface, dividedby εo. Strictly speaking–Gauss’s law for magnetic fields does not result in a usefulrelationship between magnetic fields and current distributions because it states thatthe flux of ~B through any closed surface is always zero, whether or not there arecurrents within the surface.

Ampere’s law is not formulated in terms of the magnetic flux, but rather in termsof the line integral of ~B around a closed path:∮

C

~B · d~

Ampere’s Law for a Long, Straight Conductor

In order to develop Ampere’s law, we revisit the magnetic field caused by a long,straight conductor carrying a current I. We found in a previous section that themagnetic field can be written as:

B =µo I

2πr

This results in circular concentric magnetic field lines centered on the conductor(see Fig. 7 above). If we take the line integral of ~B around a circle with radius r,we find the following:∮

C

~B · d~ =

∮B‖ d` = B

∮d` =

µoI

2πr(2πr) = µo I

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Figure 12: This figure shows three integration paths for the line integral of ~B in the vicinity ofa long, straight conductor carrying current I “out” of the plane of the page. The wire is viewedend-on.

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Figure 13: This figure shows a more general integration path for the line integral of ~B around along, straight conductor carrying current I “out” of the plane of the page. The wire is viewedend-on.

Ampere’s Law: General Statement

Suppose we have several long, straight conductros passing through the surfacebounded by the integration path C. The total magnetic field ~B at any point on thepath is the vector sum of the fields produced by the individual currents. Thus theline integral of the total ~B equals µo times the algebraic sum of the currents.

If the integration path does not enclose a particular wire, the line integral of the~B field due to that wire is zero. Thus we can replace I in our equation with Iencl,the algebraic sum of the currents enclosed by the integration path, with the sumevaluated by using the sign rule.∮

C

~B · d~ = µo Iencl (9)

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Figure 14: This figure shows the application of Ampere’s law for multiple currents. You musthave a high degree of symmetry in order to apply Ampere’s law. In this case, the currents areapproximated to be infinitely long.

Figure 15: This figure shows two long, straight conductors carrying equal currents in oppositedirections. The conductors are seen end-on, and the integration path is counterclockwise. The lineintegral

∮~B · d~ gets zero contribution from the upper and lower segments, a positive contribution

from the left segment, and a negative contribution from the right segment; the net integral is zero.

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7 Applications of Ampere’s Law

Ampere’s law is useful when we can exploit symmetry to evaluate the line integralof ~B. We illustrate this with the following examples.

Field of a Long, Straight, Current-Carrying Conducting Wire∮~B · d~ =

∮B‖ d` = B(2πr) = µoI

Field of a Long Cylindrical Conductor

Figure 16: This figure shows how the magnetic field can be calculated inside a conductor r < R.The current through the gray area is (r2/R2)I. To find the magnetic field outside the conductorr > R, we apply Ampere’s law to the circle enclosing the entire conductor.

Inside the long cylindrical conductor we find that the current enclosed is (r2/R2)I.Evaluating the line integral

∮~B · d~, we find

B =µoI

2π

r

R2(inside the conductor, r < R)

Outside the conductor, we calculate again∮~B · d~ = µoIencl, and we find the result

we calculated before:

B =µoI

2πr(outside the conductor, r > R)

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Figure 17: This figure shows the strength of the magnetic field inside and outside the conductingwire carrying a current I and having radius r = R

Field of a Solenoid

A solenoid is produced by wrapping an insulated conducting wire many times (Nturns) around a form (e.g., a cylinder or toroid) in order to confine the magneticfield to particular region of space.

Figure 18: This figure shows the magnetic field lines produced by the current in a solenoid.

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Figure 19: This figure shows how to calculate the strength of the magnetic field in a solenoid usingAmpere’s Law.

∮~B · d~ = B L

∮C

~B · d~ = B L = µoN I

B = µo n I (solenoid) (10)

where n = N/L, the number of turns per unit length.

Figure 20: This figure shows the magnitude of the magnetic field at points along the axis of asolenoid with length 4a, equal to four times its radius a. The field magnitude at each end is abouthalf its value at the center.

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Field of a Toroidal Solenoid

Figure 21: This figure shows a toroidal solenoid with only a few turns shown. In real applicationsthere are many more turns. In figure (b) there are three integration paths (black circles) used to

compute the magnetic field ~B set up by the current (shown as dots and crosses).

The figure above shows a doughnut-shaped toroidal solenoid, tightly wound withN turns of wire carrying a current I. Let’s use Ampere’s law to find the B-fieldinside and outside the toroid.

Inside the toroid the line integral (path 2) encloses N times the current I.∮C

~B · d~ = Nµo I

B · 2πr = N µo I

or

B =µoN I

2π r(toroidal solenoid) (11)

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The Bohr Magneton

In the Bohr model of the atom, electrons move in circular orbits about the nucleus.In some atoms, an external field can cause the current loops of valence electronsto become oriented preferentially with the field, so their magnetic fields add to theexternal field. We then say that the material is magnetized.

Figure 22: This figure illustrates how an electron moving with speed v in a circular orbit of radiusr has an angular moment ~L and an oppositely directed orbital magnetic dipole moment ~µ. Theelectron also has a spin angular momentum ~S and an oppositely directed spin magnetic dipolemoment ~µs (not shown in the figure).

The electron’s orbit produces both angular momentum ~L and a magnetic dipolemoment ~µ.

~L = ~r × ~p (angular momentum)

~µ = I ~A (magnetic dipole moment)

Let’s see how the magnetic dipole moment µ is related to the angular momentum L.As the electron orbits, it produces a current I = e/T = ev/2πr.

µ = I A =ev

2πrπr2 =

evr

2=

e

2m(mvr) =

e

2mL

Strictly speaking this should be a vector equation, so, we write:

~µ = − e

2m~L (12)

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If the magnetic field defines the z direction, then

µz =e

2mLz

In quantum mechanics, the angular momentum is observed to be quantized alongthe z-axis in units of h-bar where ~ = h/2π = 1.054× 10−34 J·s.

µB =e

2m(~) (Bohr magneton) (13)

where µB = 9.274× 10−24 A·m2.

Paramagnetism

Diamagnetism

Ferromagnetism

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Figure 23: This figure shows the magnetic domains produced in a single crystal of nickel. Thematerial becomes magnetized as the external magnetic field is increased. The size of the domainschange as the external magnetic field increasingly “flips” more of the dipole moments in each regionto point in the direction of ~B, thus increasing the magnetization of the material.

Figure 24: This figure shows the magnetization curve for a ferromagnetic material. The magnetiza-tion M approaches its saturation value Msat as the magnetic filed Bo (caused by external currents)becomes large.

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